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Francesco Giannino » 7.Application of differentiation: L'Hospital's Rule

L’Hospital’s Rule

Theorem. (L’Hospital’s Rule, 0/0 form). Suppose f and g are differentiable and g’(x) ≠ 0 on an open interval (a, b) containing c (except possibly at c). Suppose that
$\lim_{{x\rightarrow c}}f(x)=0, \,\,\lim_{x\rightarrow c}g(x)= 0$ $\,\, \text{and}\,\, \lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}= L, $$$
where L is a a real number, ∞, or -∞. Then
$\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=$ $\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}= L. $$$
Remark. This theorem is valid for one-sided limits as well as the two sided limit.
This theorem is also true if c = ∞, or c = -∞.

L’Hospital’s Rule

Theorem. (L’Hospital’s Rule, ∞/∞form). The previous theorem is also valid for the case when

$\lim_{x\rightarrow c}f(x)= +\infty \,\, \text{or} -\infty ,\,\,\,$ $\text{and}\,\,\lim_{x\rightarrow c}g(x)= +\infty \,\,\text{or} -\infty.$$$

Exercises

Exercise 1. Find the following limit using L’Hospital’s Rule
$\lim_{x\rightarrow 0}\frac{\sin3x}{\cos5x}.$
Solution. The assigned function yields the indeterminate form 0/0. We compute this limit applying L’Hospital’s Rule as follows:

$\lim_{x\rightarrow 0}\frac{\sin3x}{\sin5x}=$ $\lim_{x\rightarrow 0}\frac{3\cos3x}{5\cos5x}=\frac{3}{5}$$$

Exercises

Exercise 2. Find the following limit using L’Hospital’s Rule

$\lim_{x\rightarrow 0}\frac{\tan2x}{\tan3x}.$$$
Solution. The assigned function yields the indeterminate form 0/0. We compute this limit applying L’Hospital’s Rule as follows:
$\lim_{x\rightarrow 0}\frac{\tan2x}{\tan3x}$ $= \lim_{x\rightarrow 0}\frac{2\cos^{2}3x}{3\cos^{2}2x}$ $=\frac{2}{3}$$$

Exercises

Exercise 3. Find the following limit using L’Hospital’s Rule

$\lim_{x\rightarrow 0}\frac{\sin x}{x}.$

Solution. The assigned function yields the indeterminate form 0/0. We compute this limit applying L’Hospital’s Rule as follows:
$\lim_{x\rightarrow 0}\frac{\sin x}{x}=$ $\lim_{x\rightarrow 0}\frac{\cos x}{1}=1$$$

Exercises

Exercise 4. Find the following limit using L’Hospital’s Rule

$\lim_{x\rightarrow 0}\frac{1-\cos x}{x}.$
Solution. The assigned function yields the indeterminate form 0/0. We compute this limit applying L’Hospital’s Rule as follows:
$\lim_{x\rightarrow 0}\frac{1-\cos x}{x}=$ $\lim_{x\rightarrow 0}\frac{\sin x}{1} = 0$

Exercises

Exercise 5. Find the following limit using L’Hospital’s Rule

$\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}.$
Solution. The assigned function yields the indeterminate form 0/0. We compute this limit applying L’Hospital’s Rule as follows:
$\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=$ $\lim_{x\rightarrow 0}\frac{\sin x}{2x} = [\frac{0}{0}]$$$
Applying again L’Hospital’s Rule we have:
$=\lim_{x\rightarrow 0}\frac{\cos x}{2} = \frac{1}{2}$

Exercises

Exercise 6. Find the following limit using L’Hospital’s Rule

$\lim_{x\rightarrow +\infty}\frac{\exp x}{x^{2}}$
Solution. The assigned function yields the indeterminate form ∞/∞. We compute this limit applying L’Hospital’s Rule as follows:
$\lim_{x\rightarrow +\infty}\frac{\exp x}{x^{2}}=$ $\lim_{x\rightarrow +\infty}\frac{\exp x}{2x}= [\frac{+\infty}{+\infty}]$$$
Applying again L’Hospital’s Rule we have:
$=\lim_{x\rightarrow +\infty}\frac{\exp x}{2}= +\infty$

Exercises

Exercise 7. Find the following limit using L’Hospital’s Rule

$\lim_{x\rightarrow 0}x\ln x.$
Solution. The assigned function yields the indeterminate form 0·∞ when x tends to 0. Then, before applying L’Hospital’s Rule, we have to rig the assigned function to make an indeterminate quotient as follows: $f(x)=x\ln x=\frac{\ln x}{\frac{1}{x}}$
and the limit becomes: $\lim_{x\rightarrow 0}\frac{\ln x}{\frac{1}{x}}$

Exercises

Now this limit yields the indeterminate form ∞ /∞ when x tends to 0. Then, we compute this limit applying L’Hospital’s Rule as follows:
$$$ \lim_{x\rightarrow 0}\frac{\ln x}{\frac{1}{x}}=$ $\lim_{x\rightarrow 0}\frac{\frac{1}{x}}{-\frac{1}{x^{2}}}=$ $\lim_{x\rightarrow 0}\frac{1}{x}\cdot(-x^{2})=$$$

$=\lim_{x\rightarrow 0}(-x)=0.$

Exercises

Exercise 8. Find the following limit using L’Hospital’s Rule

$\lim_{x\rightarrow 0^{+}}(\frac{1}{x}-\frac{\cos 2x}{\sin 2x}).$

Solution. The assigned function yields the indeterminate form ∞ – ∞ when x tends to 0. Then, before applying L’Hospital’s Rule, we have to rig the assigned function to make an indeterminate quotient as follows:

$f(x)=\frac{1}{x}-\frac{\cos 2x}{\sin 2x}=\frac{\sin 2x-x\cos 2x}{x\sin 2x}$

and the limit becomes: $\lim_{x\rightarrow 0^{+}}\frac{\sin 2x-x\cos 2x}{x\sin 2x}$

Exercises

Now this limit yields the indeterminate form 0/0 when x tends to 0. Then, we compute this limit applying L’Hospital’s Rule as follows:
$\lim_{x\rightarrow 0^{+}}\frac{\sin 2x-x\cos 2x}{x\sin 2x}=$ $\lim_{x\rightarrow 0^{+}}\frac{2\cos 2x-\cos 2x + 2x\sin 2x}{\sin 2x + 2x\cos 2x}=$$$
$=\lim_{x\rightarrow 0^{+}}\frac{\cos 2x + 2x\sin 2x}{\sin 2x + 2x\cos 2x}=[\frac{0}{0}]$
Applying again L’Hospital’s Rule we have:
$=\lim_{x\rightarrow 0^{+}}\frac{-2\sin 2x + 2\sin 2x + 4x\cos 2x}{2\cos 2x + 2\cos 2x - 4x\sin 2x}=$
$=\lim_{x\rightarrow 0^{+}}\frac{4x\cos 2x}{4\cos 2x - 4x\sin 2x}= \frac{0}{1}= 0.$

Exercises

Exercise 9. Find the following limit using L’Hospital’s Rule

$\lim_{x\rightarrow 0^{+}}(1-2x)^{\frac{1}{x}}.$

Solution. The assigned function yields the indeterminate form 1^∞ when x tends to 0⁺. Then, before applying L’Hospital’s Rule, we have to rig the assigned function to make an indeterminate quotient as follows:

$f(x)=(1-2x)^{\frac{1}{x}}=\exp(\ln(1-2x)^{\frac{1}{x}})$

and the limit becomes: $\lim_{x\rightarrow 0^{+}}\exp(\ln(1-2x)^{\frac{1}{x}})=$
$$=\exp(\lim_{x\rightarrow 0^{+}}\ln(1-2x)^{\frac{1}{x}})$ $=\exp(\lim_{x\rightarrow 0^{+}}\frac{1}{x}\ln(1-2x))$

Exercises

where $\lim_{x\rightarrow 0^{+}}\frac{1}{x}\ln(1-2x))=[\frac{0}{0}].$

Now this limit yields the indeterminate form 0/0 when x tends to 0. So, we can compute this limit applying L’Hospital’s Rule as follows:
where
$$$\exp(\lim_{x\rightarrow 0^{+}}\frac{1}{x}\ln(1-2x))=$ $\exp(\lim_{x\rightarrow 0^{+}}\frac{\frac{-2}{1-2x}}{1})=$
$=\exp (-2)$