More on limits of functions

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

Theorem:

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

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

Proof.

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.

Corollary.

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.

Exercise.

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

$\frac{p}{q}b^{p-q}=\frac{p}{q}a^{\frac{p-q}{q}}=\frac{p}{q}a^{\frac{p}{q}-1}$

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

Proof.

(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

$(1+1/x)^{x}=(1-1/y)^{-y}=\frac{1}{(\frac{y-1}{y})^{y}}=(\frac{y}{y-1})^{y}=(1+\frac{1}{y-1})^{y-1}(1+\frac{1}{y-1})$

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.

Corollary.

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

The above is an exercise for you.

Corollary.

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

The above is an exercise for you.

Corollary.

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

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

Definition.

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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