Indeterminate forms of Limits
Examples with detailed solutions and exercises that solves limits questions related to indeterminate forms such as :
TheoremA second version of L'Hopital's rule allows us to replace the limit problem![]() with another simpler problem to solve. ![]()
Example 1Find the limit \( \lim_{x\to \infty} \dfrac{\ln x}{x} \)Solution to Example 1: Since \( \lim_{x\to \infty} \ln x = \infty\) and \( \lim_{x\to \infty} x = \infty\) we have the indeterminate form \( \lim_{x\to \infty} \dfrac{\ln x}{x} = \dfrac{\infty}{\infty} \) The above L'Hopital's rule can be used to evaluate the given limit question as follows \( \lim_{x\to \infty} \dfrac{\ln x}{x} = \lim_{x\to \infty} \dfrac{ (\ln x)' }{(x)'} \) Evaluate the derivatives in the numerator and the denominator \(= \lim_{x\to \infty} \dfrac{ 1/x }{1}\) Evaluate the limits in the numerator and denominator \( \lim_{x\to \infty} (1/x) = 0 \) and \( \lim_{x\to \infty} 1 = 1 \) We now evaluate the given limit \( \lim_{x\to \infty} \dfrac{\ln x}{x} = \lim_{x\to \infty} \dfrac{ 0 }{1} = 0 \)
Example 2Find \( \lim_{x\to \infty} x e^{-x} \)Solution to Example 2: Note that \( \lim_{x\to \infty} x = \infty \) and \( \lim_{x\to \infty} e^{-x} = 0 \) This is the indeterminate form \( \infty \cdot 0 \). The idea is to convert it into to the indeterminate form \( \dfrac{\infty}{\infty} \) and use L'Hopital's theorem. Note that \( \lim_{x\to \infty} x e^{-x} = \lim_{x\to \infty} \dfrac{ x}{ e^x} = \dfrac{ \infty }{ \infty }\) We apply the above L'Hopital's theorem \( = \lim_{x\to \infty} \dfrac{ (x)'}{ (e^x)'} = \lim_{x\to \infty} \dfrac{ 1 }{ e^x} = 0\)
Example 3Find \( \lim_{x\to \infty} ( 1 + 1/x)^x \)
Solution to Example 3:
Example 4Find the limit \( \lim_{x\to 0^+} ( 1 / x - 1 / \sin x )\)Solution to Example 4: Note that \( \lim_{x\to 0^+} ( 1 / x ) = + \infty \) and \( \lim_{x\to 0^+} ( 1 / \sin x ) = + \infty \) This limit has the indeterminate form \( \infty - \infty \) and has to be converted to another form by combining \( 1 / x - 1 / sin x \) \( \lim_{x\to 0^+} ( 1 / x - 1 / \sin x ) = \lim_{x\to 0^+} \dfrac{\sin x - x}{x \sin x} = \dfrac{0}{0} \) We now have the indeterminate form 0 / 0 and we can use the L'Hopital's theorem. \( \lim_{x\to 0^+} \dfrac{\sin x - x}{x \sin x} = \lim_{x\to 0^+} \dfrac{(\sin x - x)'}{(x \sin x)'} = \lim_{x\to 0^+} \dfrac{\cos x - 1}{ \sin x + x \cos x} = \dfrac{0}{0} \) We have again the indeterminate form 0 / 0 and use the L'Hopital's theorem one more time. \( = \lim_{x\to 0^+} \dfrac{(\cos x - 1)'}{ (\sin x + x \cos x)'} \) \( = \lim_{x\to 0^+} \dfrac{-\sin x}{ \cos x + \cos x - x \sin x } = \dfrac{0}{2} = 0\)
Example 5Find the limit \( \lim_{x\to 0^+} x^x \)Solution to Example 5: We have the indeterminate form 00. Let y = x x and ln y = ln (x x) = x ln x. Let us now find the limit of ln y \( \lim_{x\to 0^+} \ln y = \lim_{x\to 0^+} x \ln x = 0 \cdot \infty\) The above limit has the indeterminate form \( 0 \cdot \infty\). We have convert it as follows \( \lim_{x\to 0^+} x \ln x = \lim_{x\to 0^+} \dfrac{\ln x}{1/x} = \dfrac{\infty}{\infty} \) It now has the indeterminate form \( \dfrac{\infty}{\infty} \) and we can use the L'Hopital's theorem \( \lim_{x\to 0^+} \dfrac{\ln x}{1/x} = \lim_{x\to 0^+} \dfrac{(\ln x)'}{(1/x)'} \) \( = \lim_{x\to 0^+} \dfrac{1/x}{-1/x^2} \) \( = \lim_{x\to 0^+} - x = 0\) The limit of ln y = 0 and the limit of y = xx is equal to \( e^0 = 1 \) ExercisesFind the limits1. \( \lim_{x\to \infty} (\ln x)^{1/x} \) 2. \( \lim_{x\to \infty} (\ln x - \ln (1 + x)) \) 3. \( \lim_{x\to \infty} \dfrac{x}{e^x} \) 4. \( \lim_{x\to 0^+} x^{\sin x} \) Solutions to Above Exercises1. 12. 0 3. 0 4. 1 More Links on LimitsCalculus Tutorials and ProblemsLimits of Absolute Value Functions Questions |