Limit of Indeterminate Form

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

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

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

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

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

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

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

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

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

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

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

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

Now note that since

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

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

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

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 $${\lim_{x \rightarrow a + 0} \frac{f^{\prime}(x)}{g^{\prime}(x)} = l}$$ exists, then $${\lim_{x \rightarrow a + 0} \frac{f(x)}{g(x)} = l}$$.

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

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

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

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

Thus,

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

Example 2..13   Evaluate the following limit.

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

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

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

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

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

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

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

Now apply L'Hospital's Theorem, we have

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

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.

$$\lim_{x \rightarrow 0}\frac{\sin{x} - x\cos{x}}{\sin{x} - x} = \big(\frac{0}{0}\big)$$

$$\stackrel{*}{=} \lim_{x \rightarrow 0}\frac{x\sin{x}}{\cos{x} - 1} = \big(\frac{0}{0}\big)$$

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

Symbol

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

Exercise 2..13   Evaluate the following limit.

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

SOLUTION

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

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

2. $(f(x) - g(x)) \rightarrow (\infty - \infty)$

$$f(x) - g(x)$$

$$= \frac{\frac{1}{g(x)} - \frac{1}{f(x)}}{\frac{1}{f(x)g(x)}}$$

$$\rightarrow (\frac{0}{0})$$

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 $${\frac{1}{x^2} \sin{x}}$$ by $${\frac{\sin{x}}{x^2}}$$. Then it is indeterminat form of $${\frac{0}{0}}$$. Thus by L'Hospital's Theorem, we have

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

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

Example 2..14   Evaluate the following limit.

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

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

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

Therefore,

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

Exercise 2..14   Evaluate the following limit.

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

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

$$\lim_{x \rightarrow 0+}(1 + x)^{\frac{1}{x}} = \lim_{x \rightarrow 0+}e^{\frac{1}{x} \log{(1 + x)}} = \lim_{x \rightarrow 0+}e^{\frac{\log{(1+x)}}{x}}$$

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

Exercise A

1.

Find the limit of the following functions:

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

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

Exercise B

1.

Find the limit of the following functions:

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

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

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