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:
\[\lim\limits_{x \to \pi /2} \ln y = \lim\limits_{x \to \pi /2} \left( {\tan x\ln \sin x} \right) = \lim\limits_{x \to \pi /2} \frac{{\ln \sin x}}{{\cot x}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \pi /2} \frac{{{{\left( {\ln \sin x} \right)}^\prime }}}{{{{\left( {\cot x} \right)}^\prime }}} = \lim\limits_{x \to \pi /2} \frac{{\frac{1}{{\sin x}} \cdot \cos x}}{{ - \frac{1}{{{{\sin }^2}x}}}} = - \lim\limits_{x \to \pi /2} \frac{{{{\sin }^2}x \cdot \cos x}}{{\sin x}} = - \lim\limits_{x \to \pi /2} \left( {\sin x\cos x} \right) = - \lim\limits_{x \to \pi /2} \sin x \cdot \lim\limits_{x \to \pi /2} \cos x = - 1 \cdot 0 = 0.\]
Then the final answer is
\[\lim\limits_{x \to \pi /2} y = {e^0} = 1.\]
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,
\[\lim\limits_{x \to \pi /2} y = {e^0} = 1.\]
