# L’Hopital’s Rule

## Solved Problems

### Example 5.

Calculate the limit $\lim\limits_{t \to \infty } \sqrt[t]{{{t^3}}}.$

Solution.

In this example we deal with an indeterminate form of type $$\infty^0.$$ Let $$y = \sqrt[t]{{{t^3}}}.$$ After taking logarithms of both sides, we have

$\ln y = \ln \sqrt[t]{{{t^3}}} = \ln {\left( {{t^3}} \right)^{1/t}} = {\ln {t^{\frac{3}{t}}} }= \frac{3}{t}\ln t = \frac{{3\ln t}}{t}.$

Now we apply L'Hopital's rule:

$\lim\limits_{t \to \infty } \ln y = \lim\limits_{t \to \infty } \frac{{3\ln t}}{t} = \left[ {\frac{\infty }{\infty }} \right] = 3\lim\limits_{t \to \infty } \frac{{{{\left( {\ln t} \right)}^\prime }}}{{t'}} = 3\lim\limits_{t \to \infty } \frac{{1/t}}{1} = 3 \cdot 0 = 0.$

Then

$\lim\limits_{t \to \infty } y = {e^0} = 1.$

### Example 6.

Find the limit $\lim\limits_{x \to 1} {x^{\frac{1}{{1 - x}}}}.$

Solution.

We have an indeterminate form of type $$1^{\infty}.$$ Let $$y = {x^{\frac{1}{{1 - x}}}}.$$ Then

$\ln y = \ln {x^{\frac{1}{{1 - x}}}} = \frac{{\ln x}}{{1 - x}}.$

Using L'Hopital's rule, we get

$\lim\limits_{x \to 1} \ln y = \mathop {\lim }\limits_{x \to 1} \frac{{\ln x}}{{1 - x}} = \lim\limits_{x \to 1} \frac{{{{\left( {\ln x} \right)}^\prime }}}{{{{\left( {1 - x} \right)}^\prime }}} = \lim\limits_{x \to 1} \frac{{1/x}}{{ - 1}} = - \lim\limits_{x \to 1} \frac{1}{x} = - 1.$

Hence,

$\lim\limits_{x \to 1} y = {e^{ - 1}} = \frac{1}{e}.$

### Example 7.

Calculate the limit $\lim\limits_{x \to 0 + } {x^x}.$

Solution.

Let $$y = x^x.$$ Take logarithms of both sides:

$\ln y = \ln {x^x} = x\ln x.$

Find the limit of $$\ln y:$$

$\lim\limits_{x \to 0 + } \ln y = \lim\limits_{x \to 0 + } \left( {x\ln x} \right) = \lim\limits_{x \to 0 + } \frac{{\ln x}}{{1/x}} = \left[ { - \frac{\infty }{\infty }} \right] = \lim\limits_{x \to 0 + } \frac{{{{\left( {\ln x} \right)}^\prime }}}{{{{\left( {1/x} \right)}^\prime }}} = \lim\limits_{x \to 0 + } \frac{{1/x}}{{ - 1/{x^2}}} = - \lim\limits_{x \to 0 + } x = 0.$

Hence,

$\lim\limits_{x \to 0 + } y = {e^0} = 1.$

### Example 8.

Calculate the limit $\lim\limits_{x \to 0 + } {\left( {\arcsin x} \right)^{2x}}.$

Solution.

We have an indeterminate form of type $$0^0.$$ Let $$y = {\left( {\arcsin x} \right)^{2x}}.$$ Then after taking logarithms we have

$\ln y = \ln {\left( {\arcsin x} \right)^{2x}} = 2x\ln \left( {\arcsin x} \right).$

Apply L'Hopital's rule twice to get

$\lim\limits_{x \to 0 + } \ln y = \lim\limits_{x \to 0 + } \left[ {2x\ln \left( {\arcsin x} \right)} \right] = 2\lim\limits_{x \to 0 + } \frac{{\ln \left( {\arcsin x} \right)}}{{1/x}} = \left[ {\frac{\infty }{\infty }} \right] = 2\lim\limits_{x \to 0 + } \frac{{{{\left( {\ln \left( {\arcsin x} \right)} \right)}^\prime }}}{{{{\left( {1/x} \right)}^\prime }}} = 2\lim\limits_{x \to 0 + } \frac{{\frac{1}{{\arcsin x}} \cdot \frac{1}{{\sqrt {1 - {x^2}} }}}}{{ - 1/{x^2}}} = - 2\lim\limits_{x \to 0 + } \frac{{{x^2}}}{{\arcsin x}} \cdot \lim\limits_{x \to 0 + } \frac{1}{{\sqrt {1 - {x^2}} }} = - 2\lim\limits_{x \to 0 + } \frac{{{x^2}}}{{\arcsin x}} \cdot 1 = \left[ {\frac{0}{0}} \right] = - 2\lim\limits_{x \to 0 + } \frac{{{{\left( {{x^2}} \right)}^\prime }}}{{{{\left( {\arcsin x} \right)}^\prime }}} = - 2\lim\limits_{x \to 0 + } \frac{{2x}}{{\frac{1}{{\sqrt {1 - {x^2}} }}}} = - 4\lim\limits_{x \to 0 + } \left( {x\sqrt {1 - {x^2}} } \right) = - 4 \cdot 0 = 0.$

Thus,

$\lim\limits_{x \to 0 + } y = {e^0} = 1.$

### Example 9.

Find the limit $\lim\limits_{x \to \pi /2} {\left( {\sin x} \right)^{\tan x}}.$

Solution.

Direct substitution leads to the indeterminate form of type $$1^{\infty}.$$ Let $$y = {\left( {\sin x} \right)^{\tan x}}.$$ Take logarithms of both sides.

$\ln y = \ln {\left( {\sin x} \right)^{\tan x}} = \tan x\ln \sin x.$

Apply L'Hopital's rule:

$\lim\limits_{x \to \pi /2} \ln y = \lim\limits_{x \to \pi /2} \left( {\tan x\ln \sin x} \right) = \lim\limits_{x \to \pi /2} \frac{{\ln \sin x}}{{\cot x}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \pi /2} \frac{{{{\left( {\ln \sin x} \right)}^\prime }}}{{{{\left( {\cot x} \right)}^\prime }}} = \lim\limits_{x \to \pi /2} \frac{{\frac{1}{{\sin x}} \cdot \cos x}}{{ - \frac{1}{{{{\sin }^2}x}}}} = - \lim\limits_{x \to \pi /2} \frac{{{{\sin }^2}x \cdot \cos x}}{{\sin x}} = - \lim\limits_{x \to \pi /2} \left( {\sin x\cos x} \right) = - \lim\limits_{x \to \pi /2} \sin x \cdot \lim\limits_{x \to \pi /2} \cos x = - 1 \cdot 0 = 0.$

$\lim\limits_{x \to \pi /2} y = {e^0} = 1.$

### Example 10.

Calculate the limit $\lim\limits_{x \to + \infty } {\frac{{{x^n}}}{{{e^x}}}}.$

Solution.

The function has an indeterminate form of type $$\frac{\infty}{\infty}.$$ Apply L'Hopital's rule $$n$$ times:

$\lim\limits_{x \to + \infty } \frac{{{x^n}}}{{{e^x}}} = \lim\limits_{x \to + \infty } \frac{{{{\left( {{x^n}} \right)}^\prime }}}{{{{\left( {{e^x}} \right)}^\prime }}} = \lim\limits_{x \to + \infty } \frac{{n{x^{n - 1}}}}{{{e^x}}} = \lim\limits_{x \to + \infty } \frac{{{{\left( {n{x^{n - 1}}} \right)}^\prime }}}{{{{\left( {{e^x}} \right)}^\prime }}} = \lim\limits_{x \to + \infty } \frac{{n\left( {n - 1} \right){x^{n - 2}}}}{{{e^x}}} = \ldots = \lim\limits_{x \to + \infty } \frac{{n\left( {n - 1} \right)\left( {n - 1} \right) \ldots 1}}{{{e^x}}} = \lim\limits_{x \to + \infty } \frac{{n!}}{{{e^x}}} = n!\lim\limits_{x \to + \infty } \frac{1}{{{e^x}}} = n! \cdot 0 = 0.$

### Example 11.

Find the limit $\lim\limits_{x \to \pi /2} {\frac{{\tan x}}{{\tan 3x}}}.$

Solution.

According to L'Hopital's rule, we differentiate both the numerator and denominator a few times until the indeterminate form disappears:

$\lim\limits_{x \to \pi /2} \frac{{\tan x}}{{\tan 3x}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \pi /2} \frac{{{{\left( {\tan x} \right)}^\prime }}}{{{{\left( {\tan 3x} \right)}^\prime }}} = \lim\limits_{x \to \pi /2} \frac{{{{\sec }^2}x}}{{3{{\sec }^2}3x}} = \frac{1}{3}\lim\limits_{x \to \pi /2} \frac{{\frac{1}{{{{\cos }^2}x}}}}{{\frac{1}{{{{\cos }^2}3x}}}} = \frac{1}{3}\lim\limits_{x \to \pi /2} \frac{{{{\cos }^2}3x}}{{{{\cos }^2}x}} = \left[ {\frac{0}{0}} \right] = \frac{1}{3}\lim\limits_{x \to \pi /2} \frac{{{{\left( {{{\cos }^2}3x} \right)}^\prime }}}{{{{\left( {{{\cos }^2}x} \right)}^\prime }}} = \frac{1}{3}\lim\limits_{x \to \pi /2} \frac{{2\cos 3x \cdot \left( { - 3\sin 3x} \right)}}{{2\cos x \cdot \left( { - \sin x} \right)}} = \lim\limits_{x \to \pi /2} \frac{{\cos 3x\sin 3x}}{{\cos x\sin x}} = \lim\limits_{x \to \pi /2} \frac{{\cos 3x}}{{\cos x}} \cdot \lim\limits_{x \to \pi /2} \frac{{\sin 3x}}{{\sin x}} = \lim\limits_{x \to \pi /2} \frac{{\cos 3x}}{{\cos x}} \cdot \frac{{\sin \frac{{3\pi }}{2}}}{{\sin \frac{\pi }{2}}} = \lim\limits_{x \to \pi /2} \frac{{\cos 3x}}{{\cos x}} \cdot \frac{{\left( { - 1} \right)}}{1} = - \lim\limits_{x \to \pi /2} \frac{{\cos 3x}}{{\cos x}} = \left[ {\frac{0}{0}} \right] = - \lim\limits_{x \to \pi /2} \frac{{{{\left( {\cos 3x} \right)}^\prime }}}{{{{\left( {\cos x} \right)}^\prime }}} = - \lim\limits_{x \to \pi /2} \frac{{\left( { - 3\sin 3x} \right)}}{{\left( { - \sin x} \right)}} = - 3\lim\limits_{x \to \pi /2} \frac{{\sin 3x}}{{\sin x}} = - 3 \cdot \frac{{\left( { - 1} \right)}}{1} = 3.$

### Example 12.

Calculate the limit $\lim\limits_{x \to \pi /2} {\left( {\tan x} \right)^{\cos x}}.$

Solution.

Here we deal with an indeterminate form of type $$\infty^0.$$ Let $$y = {\left( {\tan x} \right)^{\cos x}}.$$ Then

$\ln y = \ln {\left( {\tan x} \right)^{\cos x}} = \cos x\ln \tan x.$

Then the limit becomes

$L = \lim\limits_{x \to \pi /2} \ln y = \lim\limits_{x \to \pi /2} \left( {\cos x\ln \tan x} \right) = \lim\limits_{x \to \pi /2} \frac{{\ln \tan x}}{{\sec x}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \pi /2} \frac{{{{\left( {\ln \tan x} \right)}^\prime }}}{{{{\left( {\sec x} \right)}^\prime }}} = \lim\limits_{x \to \pi /2} \frac{{\frac{1}{{\tan x}} \cdot {{\sec }^2}x}}{{\sec x \cdot \tan x}} = \lim\limits_{x \to \pi /2} \frac{{\sec x}}{{{{\tan }^2}x}} = \left[ {\frac{\infty }{\infty }} \right].$

As it can be seen, we still have an indeterminate form, so we differentiate the numerator and denominator one more time:

$L = \lim\limits_{x \to \pi /2} \frac{{{{\left( {\sec x} \right)}^\prime }}}{{{{\left( {{{\tan }^2}x} \right)}^\prime }}} = \lim\limits_{x \to \pi /2} \frac{{\sec x\tan x}}{{\tan x{{\sec }^2}x}} = \frac{1}{2}\lim\limits_{x \to \pi /2} \frac{1}{{\sec x}} = \frac{1}{2}\lim\limits_{x \to \pi /2} \cos x = \frac{1}{2} \cdot 0 = 0.$

Hence,

$\lim\limits_{x \to \pi /2} y = {e^0} = 1.$