Find the limit as x approaches infinity of x^(3/x).

Question:

Find {eq}\; \lim_{x\rightarrow \, \infty} x^{(3/x)} {/eq}.

L'Hopital's Rule:


L'Hopital's rule is a method to evaluate the limit.

L'Hopital rule is applied to evaluate the limit of the indetermined form {eq}\dfrac{0}{0}, \dfrac{\infty}{\infty} {/eq}

It is defined as:

{eq}\lim_{x \rightarrow a}\dfrac{f(x)}{g(x)}=\lim_{x \rightarrow a}\dfrac{f'(x)}{g'(x)} {/eq}

Answer and Explanation:


Given

{eq}\lim _{x\rightarrow \infty }x^{\dfrac{3}{x}} {/eq}


We have to evaluate the limit.

{eq}\begin{align} \lim _{x\rightarrow \infty }x^{\dfrac{3}{x}} &=\lim _{x\rightarrow \infty }e^{\ln (x)^{\dfrac{3}{x}}}\\ \\ &=\lim _{x\rightarrow \infty }e^{\dfrac{3}{x}\ln (x)}\\ \\ &=e^{3\lim _{x\rightarrow \infty }\dfrac{\ln x}{x}}\ \left [ \dfrac{\infty }{\infty } \right ]\\ \\ &=e^{3\lim _{x\rightarrow \infty }\dfrac{\dfrac{\mathrm{d} }{\mathrm{d} x}(\ln x)}{\dfrac{\mathrm{d} }{\mathrm{d} x}(x)}} \ & \left [ \text{Applying L'Hopital's Rule} \right ]\\ \\ &=e^{3\lim _{x\rightarrow \infty }\left ( \dfrac{\dfrac{1}{x}}{1} \right )}\ & \left [ \because \dfrac{\mathrm{d} }{\mathrm{d} x}(\log (x))=\dfrac{1}{x}, \dfrac{\mathrm{d} }{\mathrm{d} x}(x)=1 \right ]\\ \\ &=e^{3\lim _{x\rightarrow \infty }\dfrac{1}{x}}\\ \\ &=e^{3\left ( \dfrac{1}{\infty} \right )}\\ \\ &=e^{3(0)}\\ \\ &=e^{0}\\ \\ &=1 \end{align} {/eq}


{eq}\color{blue}{\lim _{x\rightarrow \infty }x^{\dfrac{3}{x}}=1} {/eq}


Learn more about this topic:

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How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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