Category Archives: calculus

Applications of Derivatives IITJEE Maths tutorial: practice problems part IV

Question 1.

If the point on y = x \tan {\alpha} - \frac{ax^{2}}{2u^{2}\cos^{2}{\alpha}}, where \alpha>0, where the tangent is parallel to y=x has an ordinate \frac{u^{2}}{4a}, then what is the value of \alpha?

Question 2:

Prove that the segment of the tangent to the curve y=c/x, which is contained between the coordinate axes is bisected at the point of tangency.

Question 3:

Find all the tangents to the curve y = \cos{(x+y)} for -\pi \leq x \leq \pi that are parallel to the line x+2y=0.

Question 4:

Prove that the curves y=f(x), where f(x)>0, and y=f(x)\sin{x}, where f(x) is a differentiable function have common tangents at common points.

Question 5:

Find the condition that the lines x \cos{\alpha} + y \sin{\alpha} = p may touch the curve (\frac{x}{a})^{m} + (\frac{y}{b})^{m}=1.

Question 6:

Find the equation of a straight line which is tangent to one point and normal to the point on the curve y=8t^{3}-1, and x=4t^{2}+3.

Question 7:

Three normals are drawn from the point (c,0) to the curve y^{2}=x. Show that c must be greater than 1/2. One normal is always the x-axis. Find c for which the two other normals are perpendicular to each other.

Question 8:

If p_{1} and p_{2} are lengths of the perpendiculars from origin on the tangent and normal to the curve x^{2/3} + y^{2/3}=a^{2/3} respectively, prove that 4p_{1}^{2} + p_{2}^{2}=a^{2}.

Question 9:

Show that the curve x=1-3t^{2}, and y=t-3t^{3} is symmetrical about x-axis and has no real points for x>1. If the tangent at the point t is inclined at an angle \psi to OX, prove that 3t= \tan {\psi} +\sec {\psi}. If the tangent at P(-2,2) meets the curve again at Q, prove that the tangents at P and Q are at right angles.

Question 10:

Find the condition that the curves ax^{2}+by^{2}=1 and a^{'}x^{2} + b^{'}y^{2}=1 intersect orthogonality and hence show that the curves \frac{x^{2}}{(a^{2}+b_{1})} + \frac{y^{2}}{(b^{2}+b_{1})} = 1 and \frac{x^{2}}{a^{2}+b_{2}} + \frac{y^{2}}{(b^{2}+b_{2})} =1 also intersect orthogonally.

More later,

Nalin Pithwa.

Applications of Derivatives: Tutorial: IITJEE Maths: Part II

Another set of “easy to moderately difficult” questions:

  1. The function y = \frac{}x{1+x^{2}} decreases in the interval (a) (-1,1) (b) [1, \infty) (c) (-\infty, -1] (d) (-\infty, \infty). There are more than one correct choices. Which are those?
  2. The function f(x) = \arctan (x) - x decreases in the interval (a) (1,\infty) (b) (-1, \infty) (c) (-\infty, -\infty) (d) (0, \infty). There is more than one correct choice. Which are those?
  3. For x>1, y = \log(x) satisfies the inequality: (a) x-1>y (b) x^{2}-1>y (c) y>x-1 (d) \frac{x-1}{x}<y. There is more than one correct choice. Which are those?
  4. Suppose f^{'}(x) exists for each x and h(x) = f(x) - (f(x))^{2} + (f(x))^{3} for every real number x. Then, (a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) nothing can be said in general. Find the correct choice(s).
  5. If f(x)=3x^{2}+12x-1, when -1 \leq x \leq 2, and f(x)=37-x, when 2<x\leq 3. Then, (a) f(x) is increasing on [-1,2] (b) f(x) is continuous on [-1,3] (c) f^{'}(2) doesn’t exist (d) f(x) has the maximum value at x=2. Find all the correct choice(s).
  6. In which interval does the function y=\frac{x}{\log(x)} increase?
  7. Which is the larger of the functions \sin(x) + \tan(x) and f(x)=2x in the interval (0<x<\pi/2)?
  8. Find the set of all x for which \log {(1+x)} \leq x.
  9. Let f(x) = |x-1| + a, if x \leq 1; and, f(x)=2x+3, if x>1. If f(x) has local minimum at x=1, then a \leq ?
  10. There are exactly two distinct linear functions (find them), such that they map [-1,1] and [0,2].

more later, cheers,

Nalin Pithwa.

Applications of Derivatives: Tutorial Set 1: IITJEE Mains Maths

“Easy” questions:

Question 1:

Find the slope of the tangent to the curve represented by the curve x=t^{2}+3t-8 and y=2t^{2}-2t-5 at the point (2,-1).

Question 2:

Find the co-ordinates of the point P on the curve y^{2}=2x^{3}, the tangent at which is perpendicular to the line 4x-3y+2=0.

Question 3:

Find the co-ordinates of the point P(x,y) lying in the first quadrant on the ellipse x^{2}/8 + y^{2}/18=1 so that the area of the triangle formed by the tangent at P and the co-ordinate axes is the smallest.

Question 4:

The function f(x) = \frac{\log (\pi+x)}{\log (e+x)}, where x \geq 0 is

(a) increasing on (-\infty, \infty)

(b) decreasing on [0, \infty)

(c) increasing on [0, \pi/e) and decreasing on [\pi/e, \infty)

(d) decreasing on [0, \pi/e) and increasing on [\pi/e, \infty).

Fill in the correct multiple choice. Only one of the choices is correct.

Question 5:

Find the length of a longest interval in which the function 3\sin(x) -4\sin^{3}(x) is increasing.

Question 6:

Let f(x)=x e^{x(1-x)}, then f(x) is

(a) increasing on [-1/2, 1]

(b) decreasing on \Re

(c) increasing on \Re

(d) decreasing on [-1/2, 1].

Fill in the correct choice above. Only one choice holds true.

Question 7:

Consider the following statements S and R:

S: Both \sin(x) and \cos (x) are decreasing functions in the interval (\pi/2, \pi).

R: If a differentiable function decreases in the interval (a,b), then its derivative also decreases in (a,b).

Which of the following is true?

(i) Both S and R are wrong.

(ii) Both S and R are correct, but R is not the correct explanation for S.

(iii) S is correct and R is the correct explanation for S.

(iv) S is correct and R is wrong.

Indicate the correct choice. Only one choice is correct.

Question 8:

For which of the following functions on [0,1], the Lagrange’s Mean Value theorem is not applicable:

(i) f(x) = 1/2 -x, when x<1/2; and f(x) = (1/2-x)^{2}, when x \geq 1/2.

(ii) f(x) = \frac{\sin(x)}{x}, when x \neq 0; and f(x)=1, when x=0.

(iii) f(x)=x |x|

(iv) f(x)=|x|.

Only one choice is correct. Which one?

Question 9:

How many real roots does the equation e^{x-1}+x-2=0 have?

Question 10:

What is the difference between the greatest and least values of the function f(x) = \cos(x) + \frac{1}{2}\cos(2x) -\frac{1}{3}\cos(3x)?

More later,

Nalin Pithwa.

Applications of Derivatives: A Quick Review

Section I:

The Derivative as a Rate of Change

In case of a linear function y=mx+c, the graph is a straight line and the slope m measures the steepness of the line by giving the rate of climb of the line, the rate of change of y with respect to x.

As x changes from x_{0} to x_{1}, y changes m times as much:

y_{1}-y_{0}=m(x_{1}-x_{0})

Thus, the slope m=(y_{1}-y_{0})(x_{1}-x_{0}) gives the change in y per unit change in x.

In more general case of differentiable function y=f(x), the difference quotient

\frac{f(x+h)-f(x)}{x+h-x} = \frac{f(x+h)-f(x)}{h}, where h \neq 0

give the average rate of change of y (or f) with respect to x. The limit as h approaches zero is the derivative dy/dx = f^{'}(x), which can be interpreted as the instantaneous rate of change of f with respect to x. Since, the graph is a curve, the rate of change of y can vary from point to point.

Velocity and Acceleration:

Suppose that an object is moving along a straight line and that, for each time t during a certain time interval, the object has location/position x(t). Then, at time t+h the position of the object is x(t+h) and x(t+h)-x(t) is the change in position that the object experienced during the time period t to t+h. The ratio

\frac{x(t+h)-x(t)}{t+h-t} = \frac{x(t+h)-x(t)}{h}

gives the average velocity of the object during this time period. If

\lim_{h \rightarrow 0} \frac{x(t+h)-x(t)}{h}=x^{'}(t)

exists, then x^{'}(t) gives the instantaneous rate of change of position with respect to time. This rate of change of position is called the velocity of the object. If the velocity function is itself differentiable, then its rate of change with respect to time is called the acceleration; in symbols,

a(t) = v^{'}(t) = x^{''}(t)

The speed is by definition the absolute value of the velocity: speed at time t is |v(t)|

If the velocity and acceleration have the same sign, then the object is speeding up, but if the velocity and acceleration have opposite signs, then the object is slowing down.

A sudden change in acceleration is called a jerk. Jerk is the derivative of acceleration. If a body’s position at the time t is x(t), the body’s jerk at time t is

j = \frac{da}{dt} = \frac{d^{3}x}{dt^{3}}

Differentials

Let y = f(x) be a differentiable function. Let h \neq 0. The difference f(x+h) - f(x) is called the increment of f from x to x+h, and is denoted by \Delta f.

\Delta f = f(x+h) - f(x)

The product f^{'}(x)h is called the differential of f at x with increment h, and is denoted by df

df = f^{'}(x)h

The change in f from x to x+h can be approximated by f^{'}(x)h:

f(x+h) - f(x) = f^{'}(x)h

Tangent and Normal

Let y = f(x) be the equation of a curve, and let P(x_{0}, y_{0}) be a point on it. Let PT be the tangent, PN the normal and PM the perpendicular to the x-axis.

The slope of the tangent to the curve y = f(x) at P is given by (\frac{dy}{dx})_{(x_{0}, y_{0})}

Thus, the equation of the tangent to the curve y = f(x) at (x_{0}, y_{0}) is y - y_{0} = (\frac{dy}{dx})_{(x_{0}, y_{0})}(x-x_{0})

Since PM is perpendicular to PT, it follows that if (\frac{dy}{dx})_{(x_{0}, y_{0})} \neq 0, the slope of PN is

- \frac{1}{(\frac{dy}{dx})_{(x_{0}, y_{0})}} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}

Hence, the equation of the normal to the curve y = f(x) at (x_{0}, y_{0}) is

y - y_{0} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}(x-x_{0})

The equation of the normal parallel to the x-axis is y = y_{0}, that is, when (\frac{dy}{dx})_{(x_{0}, y_{0})} = 0. The length of the tangent at (x_{0}, y_{0}) is PT, and it is equal to

y_{0}\csc{\theta} = y_{0}\sqrt{1+\cot^{2}{\theta}} = y_{0}\sqrt{1+[(\frac{dx}{dy})_{(x_{0}, y_{0})}]^{2}}

The length of the normal is PN and it is equal to y_{0}\sec {\theta} = y_{0}\sqrt{1 + [(\frac{dy}{dx})_{(x_{0}, y_{0})}]^{2}}

If the curve is represented by x = f(t) and y = g(t), that is, parametric equations in t, then

\frac{dy}{dx} = \frac{g^{'}(t)}{f^{'}(t)} where g^{'}(t)= \frac{dy}{dt} and f^{'}(t) = \frac{dx}{dt}. In this case, the equations of the tangent and the normal are given by

y - g(t) = \frac{g^{'}(t)}{f^{'}(t)}[x - f(t)] and [y-g(t)] g^{'}(t) + [x-f(t)]f^{'}(t) = 0 respectively.

The Angle between Two Curves

The angle of intersection of two curves is defined as the angle between the two tangents at the point of intersection. Let y = f(x) and y=g(x) be two curves, and let P(x_{0}, y_{0}) be their point of intersection. Also, let \psi and \phi be the angles of inclination of the two tangents with the x-axis, and let \theta be the angle between the two tangents. Then,

\tan {\theta} = \frac{\tan{\phi}-\tan{\psi}}{1+\tan{\phi}\tan{\psi}} = \frac{g^{'}(x) - f^{'}(x)}{1+f^{'}(x)g^{'}(x)}

Example 1:

Write down the equations of the tangent and the normal to the curve y = x^{3} - 3x + 2 at the point (2,4).

Solution 1:

\frac{dy}{dx} = 3x^{2}-3 \Longrightarrow \frac{dy}{dx}_{(2,4)} = 3.4 - 3 = 9.

Hence, the equation of the tangent at (2,4) is given by y-4 = 9(x-2) \Longrightarrow 9x-y-14=0 and the equation of the normal is y - 4 = (-1/9)(x-2) \Longrightarrow x+9y -38=0.

Rolle’s Theorem and Lagrange’s Theorem:

Rolle’s Theorem:

Let f(x) be a function defined on a closed interval [a,b] such that (i) f(x) is continuous on [a,b], (ii) f(x) is derivable on (a,b), and (iii) f(a) = f(b). Then, there exists a c \in (a,b) such that f^{'}(x)=0.

For details, the very beautiful, lucid, accessible explanation in Wikipedia:

https://en.wikipedia.org/wiki/Rolle%27s_theorem

Lagrange’s theorem:

Let f(x) be a function defined on a closed interval [a,b] such that (i) f(x) is continuous on [a,b], and (ii) f(x) is derivable on (a,b). Then, there exists a c \in [a,b] such that

f^{'}(c) = \frac{f(b)-f(a)}{b-a}

Example 2:

The function f(x) = \log {\sin(x)} satisfies the conditions of Rolle’s theorem on the interval [\frac{\pi}{6}, \frac{5\pi}{6}], as the logarithmic function and \sin (x) are continuous and differentiable functions and \log {\sin (\frac{5\pi}{6})} = \log {\sin (\pi - \frac{\pi}{6})} = \log{\sin{(\frac{\pi}{6})}}.

The conclusion of Rolle’s theorem is given at c=\frac{\pi}{2}, for which f^{'}(c) = \cot (c) = \cot (\pi/2) =0.

Rolle’s theorem for polynomials:

If \phi(x) is any polynomial, then between any pair of roots of \phi(x)=0 lies a root of \phi^{'}(x)=0.

Monotonicity:

A function f(x) defined on a set D is said to be non-decreasing, increasing, non-increasing and decreasing respectively, if for any x_{1}, x_{2} \in D and x_{1} < x_{2}, we have f(x_{1}) \leq f(x_{2}), f(x_{1}) < f(x_{2}), f(x_{1}) \geq f(x_{2}) and f(x_{1}) > f(x_{2}) respectively. The function f(x) is said to be monotonic if it possesses any of these properties.

For example, f(x) = e^{x} is an increasing function, and f(x)=\frac{1}{x} is a decreasing function.

Testing monotonicity:

Let f(x) be continuous on [a,b] and differentiable on (a,b). Then,

(i) for f(x) to be non-decreasing (non-increasing) on [a,b] it is necessary and sufficient that f^{'}(x) \geq 0 (f^{'}(x) \leq 0) for all x \in (a,b).

(ii) for f(x) to be increasing (decreasing) on [a,b] it is sufficient that f^{'}(x)>0 (f^{'}(x)<0) for all x \in (a,b).

(iii) If f^{'}(x)=0 for all x in (a,b), then f is constant on [a,b].

Example 3:

Determine the intervals of increase and decrease for the function f(x)=x^{3}+2x-5.

Solution 3:

We have f^{'}(x) = 3x^{2}+2, and for any value of x, 3x^{2}+2>0. Hence, f is increasing on (-\infty, -\infty). QED.

The following is a simple criterion for determining the sign of f^{'}(x):

If a,b \geq 0, then (x-a)(x-b)>0 iff x > \max (a,b) or x < \min(a,b);

(x-a)(x-b)<0 if and only if \min(a,b) < x < \max(a,b)

Maxima and Minima:

A function has a local maximum at the point x_{0} if the value of the function f(x) at that point is greater than its values at all points other than x_{0} of a certain interval containing the point x_{0}. In other words, a function f(x) has a maximum at x_{0} if it is possible to find an interval (\alpha, \beta) containing x_{0}, that is, with \alpha < x_{0} < \beta, such that for all points different from x_{0} in (\alpha, \beta), we have f(x) < f(x_{0}).

A function f(x) has a local minimum at x_{0} if there exists an interval (\alpha, \beta) containing x_{0} such that f(x) > f(x_{0}) for x \in (\alpha, \beta) and x \neq x_{0}.

One should not confuse the local maximum and local minimum of a function with its largest and smallest values over a given interval. The local maximum of a function is the largest value only in comparison to the values it has at all points sufficiently close to the point of local maximum. Similarly, the local minimum is the smallest value only in comparison to the values of the function at all points sufficiently close to the local minimum point.

The general term for the maximum and minimum of a function is extremum, or the extreme values of the function. A necessary condition for the existence of an extremum at the point x_{0} of the function f(x) is that f^{'}(x_{0})=0, or f^{'}(x_{0}) does not exist. The points at which f^{'}(x)=0 or f^{'}(x) does not exist, are called critical points.

First Derivative Test:

(i) If f^{'}(x) changes sign from positive to negative at x_{0}, that is, f^{'}(x)>0 for x < x_{0} and f^{'}(x)<0 for x > x_{0}, then the function attains a local maximum at x_{0}.

(ii) If f^{'}(x) changes sign from negative to positive at x_{0}, that is, f^{'}(x)<0 for x<x_{0}, and f^{'}(x)>0 for x > x_{0}, then the function attains a local minimum at x_{0}.

(iii) If the derivative does not change sign in moving through the point x_{0}, there is no extremum at that point.

Second Derivative Test:

Let f be twice differentiable, and let c be a root of the equation f^{'}(x)=0. Then,

(i) c is a local maximum point if f^{''}(c)<0.

(ii) c is a local minimum point if f^{''}(c)>0.

However, if f^{''}(c)=0, then the following result is applicable. Let f^{'}(c) = f^{''}(c) = \ldots = f^{n-1}(c)=0 (where f^{r} denotes the rth derivative), but f^{(n)}(c) \neq 0.

(i) If n is even and f^{(n)}(c)<0, there is a local maximum at c, while if f^{(n)}(c)>0, there is a local minimum at c.

(ii) If n is odd, there is no extremum at the point c.

Greatest/Least Value (Absolute Maximum/Absolute Minimum):

Let f be a function with domain D. Then, f has a greatest value (or absolute maximum) at a point c \in D if f^(x) \leq f(c) for all x in D and a least value (or absolute minimum) at c, if f(x) \geq f(c) for all x in D.

If f is continuous at every point of D, and D=[a,b], a closed interval, the f assumes both a greatest value M and a least value m, that is, there are x_{1}, x_{2} \in [a,b] such that f(x_{1})=M and f(x_{2})=m, and m \leq f(x) \leq M for every x \in [a,b].

Example 4:

a) y=x^{2}, with domain (-\infty, \infty). This has no greatest value; least value at x=0

b) y=x^{2} with domain [0,2]. This has greatest value at x=2 and least value at x=0.

c) y=x^{2} with domain (0,2]. This has greatest value at x=2 and no least value.

d) y=x^{2} with domain (0,2). This has no greatest value and no least value.

Some other remarks:

The greatest (least) value of continuous function f(x) on the interval [a,b] is attained either at the critical points or at the end points of the interval. To find the greatest (least) value of the function, we have to compute its values at all the critical points on the interval (a,b), and the values f(a), f(b) of the function at the end-points of the interval, and choose the greatest (least) out of the values so obtained.

We will continue with problems on applications of derivatives later,

Nalin Pithwa.

Happy “e” day via Math: thanks to PlusMaths :-)

https://plus.maths.org/content/happy-e-day?nl=0

The Early History of Calculus Problems, II: AMS feature column

http://www.ams.org/samplings/feature-column/fc-current.cgi

The Early History of Calculus Problems: AMS Feature column

http://www.ams.org/samplings/feature-column/fc-2016-05

Can anyone have fun with infinite series?

Below is list of finitely many puzzles on infinite series to keep you a bit busy !! 🙂 Note that these puzzles do have an academic flavour, especially concepts of convergence and divergence of an infinite series.

Puzzle 1: A grandmother’s vrat (fast) requires her to keep odd number of lamps of finite capacity lit in a temple at any time during 6pm to 6am the next morning. Each oil-filled lamp lasts 1 hour and it burns oil at a constant rate. She is not allowed to light any lamp after 6pm but she can light any number of lamps before 6pm and transfer oil from some to the others throughout the night while keeping odd number of lamps lit all the time. How many fully-filled oil lamps does she need to complete her vrat?

Puzzle 2: Two number theorists, bored in a chemistry lab, played a game with a large flask containing 2 liters of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that p+2 is also a prime number. Then, the first number theorist would pipette out \frac{1}{p} litres of chemical and the second \frac{1}{(p+2)} litres. How many times do they have to play this game to empty the flask completely?

Puzzle 3: How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?

Puzzle 4: Imagine a tank that can be filled with infinite taps and can be emptied with infinite drains. The taps, turned on alone, can fill the empty tank to its full capacity in 1 hour, 3 hours, 5 hours, 7 hours and so on. Likewise, the drains opened alone, can drain a full tank in 2 hours, 4 hours, 6 hours, and so on. Assume that the taps and drains are sequentially arranged in the ascending order of their filling and emptying durations.

Now, starting with an empty tank, plumber A alternately turns on a tap for 1 hour and opens the drain for 1 hour, all operations done one at a time in a sequence. His sequence, by using t_{i} for i^{th} tap and d_{j} for j^{th} drain, can be written as follows: \{ t_{1}, d_{1}, t_{2}, d_{2}, \ldots\}_{A}.

When he finishes his operation, mathematically, after using all the infinite taps and drains, he notes that the tank is filled to a certain fraction, say, n_{A}<1.

Then, plumber B turns one tap on for 1 hour and then opens two drains for 1 hour each and repeats his sequence: \{ (t_{1},d_{1},d_{2}), (t_{2},d_{3},d_{4}), (t_{3},d_{4},d_{5}) \ldots \}_{B}.

At the end of his (B’s) operation, he finds that the tank is filled to a fraction that is exactly half of what plumber A had filled, that is, 0.5n_{A}.

How is this possible even though both have turned on all taps for 1 hour and opened all drains for 1 hour, although in different sequences?

I hope u do have fun!!

-Nalin Pithwa.

Huygen’s Clock

Ref: Calculus and Analytic Geometry, G B Thomas and Finney, 9th edition.

The problem with a pendulum clock whose bob swings in a circular arc is that the frequency of the swing depends on the amplitude of the spring. The wider the swing, the longer it takes the bob to return to centre.

This does not happen if the bob can be made to swing in a cycloid. In 1673, Chritiaan Huygens (1629-1695), the Dutch mathematician, physicist and astronomer who discovered the rings of Saturn, driven by a need to make accurate determinations of longitude at sea, designed a pendulum clock whose bob would swing in a cycloid. He  hung the bob from a fine wire constrained by guards that caused it to draw up as it swung away from the centre. How were the guards shaped? They were cycloids, too.

Aufwiedersehen,

Nalin Pithwa.

Limits that arise frequently

We continue our presentation of basic stuff from Calculus and Analytic Geometry, G B Thomas and Finney, Ninth Edition. My express purpose in presenting these few proofs is to emphasize that Calculus, is not just a recipe of calculation techniques. Or, even, a bit further, math is not just about calculation. I have a feeling that such thinking nurtured/developed at a young age, (while preparing for IITJEE Math, for example) makes one razor sharp.

We verify a few famous limits.

Formula 1:

If |x|<1, \lim_{n \rightarrow \infty}x^{n}=0

We need to show that to each \in >0 there corresponds an integer N so large that |x^{n}|<\in for all n greater than N. Since \in^{1/n}\rightarrow 1, while |x|<1. there exists an integer N for which \in^{1/n}>|x|. In other words,

|x^{N}|=|x|^{N}<\in. Call this (I).

This is the integer we seek because, if |x|<1, then

|x^{n}|<|x^{N}| for all n>N. Call this (II).

Combining I and II produces |x^{n}|<\in for all n>N, concluding the proof.

Formula II:

For any number x, \lim_{n \rightarrow \infty}(1+\frac{x}{n})^{n}=e^{x}.

Let a_{n}=(1+\frac{x}{n})^{n}. Then, \ln {a_{n}}=\ln{(1+\frac{x}{n})^{n}}=n\ln{(1+\frac{x}{n})}\rightarrow x,

as we can see by the following application of l’Hopital’s rule, in which we differentiate with respect to n:

\lim_{n \rightarrow \infty}n\ln{(1+\frac{x}{n})}=\lim_{n \rightarrow \infty}\frac{\ln{(1+x/n)}}{1/n}, which in turn equals

\lim_{n \rightarrow \infty}\frac{(\frac{1}{1+x/n}).(-\frac{x}{n^{2}})}{-1/n^{2}}=\lim_{n \rightarrow \infty}\frac{x}{1+x/n}=x.

Now, let us apply the following theorem with f(x)=e^{x} to the above:

(a theorem for calculating limits of sequences) the continuous function theorem for sequences:

Let a_{n} be a sequence of real numbers. If \{a_{n}\} be a sequence of real numbers. If a_{n} \rightarrow L and if f is a function that is continu0us at L and defined at all a_{n}, then f(a_{n}) \rightarrow f(L).

So, in this particular proof, we get the following:

(1+\frac{x}{n})^{n}=a_{n}=e^{\ln{a_{n}}}\rightarrow e^{x}.

Formula 3:

For any number x, \lim_{n \rightarrow \infty}\frac{x^{n}}{n!}=0

Since -\frac{|x|^{n}}{n!} \leq \frac{x^{n}}{n!} \leq \frac{|x|^{n}}{n!},

all we need to show is that \frac{|x|^{n}}{n!} \rightarrow 0. We can then apply the Sandwich Theorem for Sequences (Let \{a_{n}\}, \{b_{n}\} and \{c_{n}\} be sequences of real numbers. if a_{n}\leq b_{n}\leq c_{n} holds for all n beyond some index N, and if \lim_{n\rightarrow \infty}a_{n}=\lim_{n\rightarrow \infty}c_{n}=L,, then \lim_{n\rightarrow \infty}b_{n}=L also) to  conclude that \frac{x^{n}}{n!} \rightarrow 0.

The first step in showing that |x|^{n}/n! \rightarrow 0 is to choose an integer M>|x|, so that (|x|/M)<1. Now, let us the rule (formula 1, mentioned above), so we conclude that:(|x|/M)^{n}\rightarrow 0. We then restrict our attention to values of n>M. For these values of n, we can write:

\frac{|x|^{n}}{n!}=\frac{|x|^{n}}{1.2 \ldots M.(M+1)(M+2)\ldots n}, where there are (n-M) factors in the expression (M+1)(M+2)\ldots n, and

the RHS in the above expression is \leq \frac{|x|^{n}}{M!M^{n-M}}=\frac{|x|^{n}M^{M}}{M!M^{n}}=\frac{M^{M}}{M!}(\frac{|x|}{M})^{n}. Thus,

0\leq \frac{|x|^{n}}{n!}\leq \frac{M^{M}}{M!}(\frac{|x|}{M})^{n}. Now, the constant \frac{M^{M}}{M!} does not change as n increases. Thus, the Sandwich theorem tells us that \frac{|x|^{n}}{n!} \rightarrow 0 because (\frac{|x|}{M})^{n}\rightarrow 0.

That’s all, folks !!

Aufwiedersehen,

Nalin Pithwa.