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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)

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Progetto "Campus Virtuale" dell'Università degli Studi di Napoli Federico II, realizzato con il cofinanziamento dell'Unione europea. Asse V - Società dell'informazione - Obiettivo Operativo 5.1 e-Government ed e-Inclusion