More on limits of functions

Before we move on to the evaluation of other limits, we prove the following useful result:


Suppose f:(a,b) \rightarrow \Re is continuous at c \in (a,b). Suppose g:I \rightarrow (a,b), when I is an open interval and x_{0} \in I. If \lim_{x \rightarrow x_{0}} g(x) exists and is equal to c, then

\lim_{x \rightarrow x_{0}}f(g(x))=f(c).


Since f is continuous at x_{0}, for each \varepsilon >0, we can find \delta^{'}>0 such that |f(y)-f(c)|<\varepsilon for |y-c|<\delta^{'}.

Since \lim_{x \rightarrow x_{0}}g(x)=c, for the given \delta^{'}, we can find a \delta>0 such that |g(x)-c|<\delta^{'} for 0<|x-x_{0}|<\delta.

So, for 0<|x-x_{0}|<\delta, we have |f(g(x))-f(c)|< \varepsilon since |g(x)-c|<\delta^{'} . QED.


Let I and J be open intervals, g:I \rightarrow J be continuous at x_{0} \rightarrow I. If f:J \rightarrow \Re is continuous at g(x_{0}) \in J, then f \circ g: I \rightarrow \Re is continuous at x_{0}. This means that the composition of continuous functions is continuous.


If f is continuous, show that |f| is continuous. (where |f|(x)=|f(x)|).

Some Useful Limits.

(1)(a) \lim_{x \rightarrow a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1} for a positive integer n. For this, note that \frac{x^{n}-a^{n}}{x-a}=x^{n-1}+x^{n-2}a+\ldots+xa^{n-2}+a^{n-1} for x \neq a, and hence,

\lim_{x \rightarrow a}\frac{x^{n}-a^{n}}{x-a}=\lim_{x \rightarrow a}(x^{n-1}+x^{n-2}a+\ldots+xa^{n-2}+a^{n-1})=na^{n-1}.

(1)(b) \lim_{x \rightarrow a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1} for a negative integer n if a \neq 0.

If n is a negative integer write n=-m, where m>0.

\lim_{x \rightarrow a}\frac{x^{n}-a^{n}}{x-a}=\lim_{x \rightarrow a}\frac{x^{-m}-a^{-m}}{x-a}=\lim_{x \rightarrow a}\frac{a^{m}-x^{m}}{x-a}.\frac{1}{x^{m}a^{m}}=-ma^{m-1}.\frac{1}{a^{2m}}=-ma^{-m-1}=na^{n-1}

(1)(c) \lim_{x \rightarrow a}\frac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a}=\frac{p}{q}a^{\frac{p}{q}-1}, for p, q \in Z and when a>0.

Write y=x^{1/q}, b=a^{1/q}. So we have have x=y^{q}, a=b^{q}.

\lim_{x \rightarrow a}\frac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a}=\lim_{y \rightarrow b}\frac{y^{p}-b^{p}}{y^{q}-b^{q}}=\lim_{y \rightarrow b}\frac{\frac{y^{p}-b^{p}}{y-b}}{\frac{y^{q}-b^{q}}{y-b}}=\frac{pb^{p-1}}{qb^{q-1}} which in turn equals


(2) \lim_{\theta \rightarrow 0}\frac{\sin{\theta}}{\theta}=1 where \theta is measured in radians.

In some cases, when the function f:\Re \rightarrow \Re is defined and there is an l such that for every \varepsilon>0 we have an x_{0} such that

|f(x)-l|<\varepsilon for x > x_{0}

we say that f(x) tends to l as x tends to \infty.We write

\lim_{x \rightarrow \infty}f(x)=l

Be warned that \infty is not a real number to which x is coming close. Similarly, we may define \lim_{x \rightarrow -\infty}f(x)=l. If for every M>0 we can find a \delta>0 such that f(x)<M for 0<|x-a|<\delta, then we can write \lim_{x \rightarrow a}f(x)=\infty. A similar definition for \lim_{x \rightarrow a}f(x)=-\infty can be given.

(3)(a) \lim_{x \rightarrow \infty}(1+1/x)^{x}=e (3b) \lim_{x \rightarrow -\infty}(1+1/x)^{x}=e


(3a) Let n \leq x < n+1 so that 1+\frac{1}{n+1}<1+1/x \leq 1+1/n giving us (1+\frac{1}{n+1})^{n}<(1+1/x)^{x} \leq (1+1/n)^{n+1}. This implies that

(1+\frac{1}{n+1})^{n+1}(1+\frac{1}{n+1})^{-1}<(1+1/x)^{x}\leq (1+1/n)^{n}(1+1/n)

Now, as x \rightarrow \infty, n also tends to \infty and in such a case both the right hand and left hand side of the above inequality tend to e. So,

\lim_{x \rightarrow \infty}(1+1/x)^{x}=e. QED.

(3b) In the case of x \rightarrow -\infty, let us write y=-x so that


Now, \lim_{y \rightarrow \infty}(1+\frac{1}{y-1})^{y-1}=e. So, we have

\lim_{x \rightarrow -\infty}(1+1/x)^{x}=\lim_{y \rightarrow \infty}(1-1/y)^{-y}=\lim (1+\frac{1}{y-1})^{y-1}(1+\frac{1}{y-1})=e. QED.


\lim_{x \rightarrow \infty}(1-1/x)^{x}=1/e.

The above is an exercise for you.


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

The above is an exercise for you.


\lim_{t \rightarrow 0}(1+t)^{1/t}=e.

It will be appropriate to mention something about left continuity and right continuity of functions.


If \lim_{x \rightarrow x_{0-}}f(x)=f(x_{0}), then we say that the function f is left continuous at x_{0}. If

\lim_{x \rightarrow x_{0+}}f(x)=f(x_{0}), then we say that the function is right continuous at x_{0}.

Thus, a function may be left continuous but not right continuous, in which case the function cannot be continuous. It may be right continuous but not left continuous, in which case also the function cannot be continuous. If a function is both left continuous and right continuous, only then is it continuous.

More later,

Nalin Pithwa

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