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

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

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

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

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

To find the limit by L'Hospital's Theorem, we usually write in the following way.

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

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

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

Exercise 2..13   Evaluate the following limit.

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

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