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Limit of Indeterminate Form

Indeterminate Form
The following cases are indeterminate.
\begin{displaymath}\begin{array}{ll} 1. & \lim_{x \to a}\frac{f(x)}{g(x)} = \lef... ...eft(\infty^0\right)\mbox{or}\ \left(0^\infty\right) \end{array}\end{displaymath}

Cauchy's Mean Value Theorem

Theorem 2..12   Let $f(x)$ and $g(x)$ be continuous on $[a,b]$ adn differentiable on $(a,b)$. If $g(a) \neq g(b)$ and $f^{\prime}(x)$ and $g^{\prime}(x)$ never takes 0 simultaneously, then there exists at leat one number $\xi \in (a,b)$ such that

$\displaystyle \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f^{\prime}(\xi)}{g^{\prime}(\xi)} $

Proof We consider a function which satisfies the conditions of Rolle's Theorem. Let

$\displaystyle G(x) = f(x) - \frac{f(b)-f(a)}{g(b)-g(a)}(g(x) - g(a)) - f(a) $

Then $G(a) = G(b) = 0$ and $G(x)$ satisfies the conditions of Rolle's Theorem. Thus there exists at least one number $\xi \in (a,b)$ such that

$\displaystyle G^{\prime}(\xi) = f^{\prime}(\xi) - \frac{f(b)-f(a)}{g(b)-g(a)}g^{\prime}(\xi) = 0. $

Now note that since

$\displaystyle f^{\prime}(\xi) - \frac{f(b)-f(a)}{g(b)-g(a)}g^{\prime}(\xi) = 0 $

if $g^{\prime}(\xi) = 0$, then $f^{\prime}(\xi) = 0$ and this violates the assumption. Therefore, $g^{\prime}(\xi) \neq 0$ and

$\displaystyle \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f^{\prime}(\xi)}{g^{\prime}(\xi)}\ensuremath{\ \blacksquare}$

L'Hospital's Theorem

Theorem 2..13   Let $f(x)$ and $g(x)$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $f(a) = g(a) = 0$ and $\displaystyle{\lim_{x \rightarrow a + 0} \frac{f^{\prime}(x)}{g^{\prime}(x)} = l}$ exists, then $\displaystyle{\lim_{x \rightarrow a + 0} \frac{f(x)}{g(x)} = l}$.

Proof Let $x$ be such that $a < x < b$ and consider

$\displaystyle \frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} $

Then by Cauchy's Mean Value Theorem, there exits at least one number $\xi \in (a,b)$ such that

$\displaystyle \frac{f(x)}{g(x)} = \frac{f^{\prime}(\xi)}{g^{\prime}(\xi)} $

Thus,

$\displaystyle \lim_{x \rightarrow a+0}\frac{f(x)}{g(x)} = \lim_{\xi \rightarrow a+0}\frac{f^{\prime}(\xi)}{g^{\prime}(\xi)} = l \ensuremath{\ \blacksquare} $

Example 2..13   Evaluate the following limit.

$\displaystyle \lim_{x \rightarrow 0}\frac{\sin{x} - x\cos{x}}{\sin{x} - x} $

SOLUTION This is indeterminate form of $\displaystyle{\frac{0}{0}}$. Then differentiate the numerator and denominator separately, we have

$\displaystyle \lim_{x \rightarrow 0}\frac{x\sin{x}}{\cos{x} - 1}. $

This is again indeterminate form of $\displaystyle{\frac{0}{0}}$. So, differentiate the numerator and denominator separately, we have

$\displaystyle \lim_{x \rightarrow 0}\frac{\sin{x} + x\cos{x}}{-\sin{x}}. $

This is again indeterminate form of $\displaystyle{\frac{0}{0}}$. So, differentiate the numerator and denominator separately, we have

$\displaystyle \lim_{x \rightarrow 0}\frac{2\cos{x} - x\sin{x}}{-\cos{x}} = \frac{2}{-1} = -2. $

Now apply L'Hospital's Theorem, we have

$\displaystyle \lim_{x \rightarrow 0}\frac{\sin{x} - x\cos{x}}{\sin{x} - x} = -2\ensuremath{\ \blacksquare}$

To find the limit by L'Hospital's Theorem, we usually write in the following way.
$\displaystyle \lim_{x \rightarrow 0}\frac{\sin{x} - x\cos{x}}{\sin{x} - x} $ $\displaystyle =$ $\displaystyle \big(\frac{0}{0}\big)$  
  $\displaystyle \stackrel{*}{=}$ $\displaystyle \lim_{x \rightarrow 0}\frac{x\sin{x}}{\cos{x} - 1} = \big(\frac{0}{0}\big)$  
  $\displaystyle \stackrel{*}{=}$ $\displaystyle \lim_{x \rightarrow 0}\frac{2\cos{x} - x\sin{x}}{-\cos{x}} = \frac{2}{-1} = -2. $  

Exercise 2..13   Evaluate the following limit.

$\displaystyle \lim_{x \rightarrow 0} \frac{1}{x^2} \sin{x}$

SOLUTION Note that this is indeterminate form of $\infty \cdot 0$. Then we replace $\displaystyle{\frac{1}{x^2} \sin{x}}$ by $\displaystyle{\frac{\sin{x}}{x^2}}$. Then it is indeterminat form of $\displaystyle{\frac{0}{0}}$. Thus by L'Hospital's Theorem, we have

$\displaystyle \lim_{x \rightarrow 0}\frac{\sin{x}}{x^2} \stackrel{*}{=} \lim_{x \rightarrow 0}\frac{\cos{x}}{2x} $

Here, $\displaystyle{ \lim_{x \rightarrow 0-}\frac{\cos{x}}{2x} = - \infty, \ \lim_{x \rightarrow 0+}\frac{\cos{x}}{2x} = \infty}$. Thus no limit exists $\ \blacksquare$

Example 2..14   Evaluate the following limit.

$\displaystyle \lim_{x \rightarrow 0+}\left(\frac{e^x}{x} - \frac{1}{x}\right)$

This is indeterminate form of . Then replace by . Then it is indeterminate form of . Thus SOLUTION$\infty - \infty$$\displaystyle{\frac{e^x}{x} - \frac{1}{x}}$$\displaystyle{\frac{e^x - 1}{x}}$$\displaystyle{\frac{0}{0}}$

$\displaystyle \lim_{x \rightarrow 0+}\frac{e^x - 1}{x} \stackrel{*}{=} \lim_{x \rightarrow 0+}\frac{e^x}{1} = 1$

Therefore,

$\displaystyle \lim_{x \rightarrow 0+}(\frac{e^x}{x} - \frac{1}{x}) = 1\ensuremath{\ \blacksquare}$

Exercise ..214   Evaluate the following limit.

$\displaystyle \lim_{x \rightarrow 0+}(1 + x)^{\frac{1}{x}}$

This is indeterminate form of . So we rewrite into . Then is indeterminate form of . Thus replace by . Then in the form of . Thus SOLUTION$\displaystyle{1^{\infty}}$$\displaystyle{(1 + x)^{\frac{1}{x}}}$$\displaystyle{e^{\frac{1}{x} \log{(1 + x)}}}$$\displaystyle{\frac{1}{x} \cdot \log{(1 + x)}}$$\infty \cdot 0$$\displaystyle{\frac{1}{x} \cdot \log{(1 + x)}}$$\displaystyle{\frac{\log{(1+x)}}{x}}$$\displaystyle{\frac{0}{0}}$
$\displaystyle \lim_{x \rightarrow 0+}(1 + x)^{\frac{1}{x}}$$\displaystyle =$$\displaystyle \lim_{x \rightarrow 0+}e^{\frac{1}{x} \log{(1 + x)}} = \lim_{x \rightarrow 0+}e^{\frac{\log{(1+x)}}{x}}$    $\displaystyle \stackrel{*}{=}$$\displaystyle \lim_{x \rightarrow 0+}e^{\frac{1/(1+x)}{1}} = e\ensuremath{\ \blacksquare}$  

Exercise A
1.
Find the limit of the following functions:

(a)$\displaystyle{\lim_{x \rightarrow 0+}\frac{\sin{x}}{\sqrt{x}}}$(b)$\displaystyle{\lim_{x \rightarrow 1}\frac{\log{x}}{1-x}}$(c)$\displaystyle{\lim_{x \rightarrow 4}\frac{\sqrt{x}-2}{x-4}}$(d)$\displaystyle{\lim_{x \rightarrow 0}\frac{2^{x} - 1}{x}}$

(e)$\displaystyle{\lim_{x \rightarrow 0}\frac{1 - \cos{x}}{3x}}$(f)$\displaystyle{\lim_{x \rightarrow \infty}\frac{x-1}{x+1}}$(g)$\displaystyle{\lim_{x \rightarrow \infty}\frac{2\sin{x}}{x}}$(h)$\displaystyle{\lim_{x \rightarrow 0}\frac{e^{x} - e^{-x}}{x}}$

Exercise B


1.
Find the limit of the following functions:

(a)$\displaystyle{\lim_{x \rightarrow 0}\frac{\sin{2x}}{\sin{3x}}}$(b)$\displaystyle{\lim_{x \rightarrow 0}\frac{\cos{x} - 1}{x}}$(c)$\displaystyle{\lim_{x \rightarrow 0}\frac{\sin^{-1}{x}}{x}}$(d)$\displaystyle{\lim_{x \rightarrow 0}x\sin{\frac{1}{x}}}$

(e)$\displaystyle{\lim_{x \rightarrow 0}(\frac{1}{x^{2}} - \frac{1}{\sin^{2}{x}})}$(f)$\displaystyle{\lim_{x \rightarrow \frac{\pi}{2}}(1 - \sin{x})^{\cos{x}}}$(g)$\displaystyle{\lim_{x \rightarrow 0}(\frac{e^{x} - 1}{x})^{1/x}}$

(h)$\displaystyle{\lim_{x \rightarrow 0}(1 - x)^{\frac{1}{\sin{x}}}}$



   
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