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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)} $

This is wrong Apply the Mean Value Theorem to $f(x)$ and $g(x)$. Then

$\displaystyle f(b) - f(a) = f^{\prime}(\xi_{1})(b - a) $

$\displaystyle g(b) - g(a) = g^{\prime}(\xi_{2})(b - a) $

Now dividing the top equation by the bottom equation, we obtain

$\displaystyle \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f^{\prime}(\xi_{1})}{g^{\prime}(\xi_{2})} $

Here, there is no guarantee that $\xi_{1}$ and $\xi_{2}$ are equal.
Proof
Check
$G(a) = f(a) - \frac{f(b)-f(a)}{g(b)-g(a)}(g(a) - g(a)) - f(a) = 0$. Then $g(b) = f(b) - \frac{f(b)-f(a)}{g(b)-g(a)}(g(b) - g(a)) - f(a) = f(b) - f(b) + f(a) - f(a) = 0$.

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 $\displaystyle{\lim_{x \rightarrow a + 0} \frac{f(x)}{g(x)} = l}$.

L'Hospital's Theorem 1. First make sure that limit is in indeterminate form of either $(\frac{0}{0})$ or $(\frac{\infty}{\infty})$.
2. Differentiate the numerator and denominator separetely.
3. After differentiation, Simplify the expression.
4. If it is indeterminate form again, repeat 2.3.

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

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

Check
$(\sin{x} + x\cos{x})' = (\sin{x})' + (x\cos{x})' = \cos{x} + x'(\cos{x}) + x(\cos{x})' = \cos{x} + \cos{x} -x\sin{x} = 2\cos{x} -x\sin{x}$.

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. $  

Symbol
In this paper, when we apply L'Hospital's rule, we use the following symbol $\stackrel{*}{=}$ .

Exercise 2..13   Evaluate the following limit.

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

SOLUTION

Other than $(\frac{0}{0}), (\frac{\infty}{\infty})$ Note that L'Hospital's theorem only can apply $\displaystyle{\left(\frac{0}{0}\right), \left(\frac{\infty}{\infty}\right)}$. Other indeterminate form appears, you must change into $\displaystyle{\left(\frac{0}{0}\right), \left(\frac{\infty}{\infty}\right)}$.
1. $f(x)g(x) \rightarrow (0\cdot \infty)$

$\displaystyle f(x)g(x) = \frac{f(x)}{\frac{1}{g(x)}} \rightarrow (\frac{0}{0})$


2. $(f(x) - g(x)) \rightarrow (\infty - \infty)$
    $\displaystyle f(x) - g(x)$  
  $\displaystyle =$ $\displaystyle \frac{\frac{1}{g(x)} - \frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}$  
  $\displaystyle \rightarrow$ $\displaystyle (\frac{0}{0})$  


3. $(f(x)^{g(x)}) \rightarrow (1^{\infty}$ or $\infty^{0})$ $f(x)^{g(x)} = e^{\log{f(x)^{g(x)}}} = e^{g(x)\log{f{(x)}}} \rightarrow (e^{\infty\cdot 0} $   or$ e^{0 \cdot \infty}) $

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.

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

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

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

Therefore,

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

Exercise 2..14   Evaluate the following limit.

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

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

$\displaystyle \lim_{x \rightarrow 0+}(1 + x)^{\frac{1}{x}}$ $\displaystyle =$ $\displaystyle \lim_{x \rightarrow 0+}e^{\frac{1}{x} \log{(1 + x)}} = \lim_{x \rightarrow 0+}e^{\frac{\log{(1+x)}}{x}}$  
  $\displaystyle \stackrel{*}{=}$ $\displaystyle \lim_{x \rightarrow 0+}e^{\frac{1/(1+x)}{1}} = e\ensuremath{ \blacksquare}$  

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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