# Thread: Limit Problem - # 2

1. ## Limit Problem - # 2

$\displaystyle \lim x \rightarrow \infty$

$\displaystyle x^{\dfrac{4}{x}}$

$\displaystyle \infty^{\dfrac{4}{\infty}}$

$\displaystyle \infty^{0}$

Quotient Rule or not - I think not.

With Quotient Rule

$\displaystyle \ln[x^{\dfrac{4}{x}}]$

$\displaystyle \dfrac{4\ln x}{x}$

$\displaystyle \dfrac{(x)(\dfrac{4}{x} - \ln x (1)}{x^{2}}$

$\displaystyle \dfrac{4 (\infty) - 4 \ln (\infty)}{\infty^{2}}$

Without Quotient

$\displaystyle \ln[x^{\dfrac{4}{x}}]$

$\displaystyle \dfrac{4 \ln x}{x}$

$\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}$

$\displaystyle \dfrac{4 \ln (\infty)}{(\infty)}$

$\displaystyle \dfrac{(\infty)}{(\infty)} =$ Indeterminate What now?

3. ## Re: Limit Problem - # 2

Originally Posted by Jason76
$\displaystyle \lim x \rightarrow \infty$

$\displaystyle x^{\dfrac{4}{x}}$

$\displaystyle \infty^{\dfrac{4}{\infty}}$

$\displaystyle \infty^{0}$

Quotient Rule or not - I think not.

With Quotient Rule

$\displaystyle \ln[x^{\dfrac{4}{x}}]$

$\displaystyle \dfrac{4\ln x}{x}$

$\displaystyle \dfrac{(x)(\dfrac{4}{x} - \ln x (1)}{x^{2}}$

$\displaystyle \dfrac{4 (\infty) - 4 \ln (\infty)}{\infty^{2}}$

Without Quotient

$\displaystyle \ln[x^{\dfrac{4}{x}}]$

$\displaystyle \dfrac{4 \ln x}{x}$

$\displaystyle \dfrac{\dfrac{d}{dx} 4 \ln x}{\dfrac{d}{dx} x}$

$\displaystyle \dfrac{4 \ln (\infty)}{(\infty)}$

$\displaystyle \dfrac{(\infty)}{(\infty)} =$ Indeterminate What now?
\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty} x^{\frac{4}{x}} &= \lim_{x \to \infty} e^{\ln{ \left[ x^{\frac{4}{x}} \right] }} \\ &= \lim_{x \to \infty} e^{ \frac{4\ln{(x)}}{x} } \\ &= e^{ \lim_{x \to \infty} \frac{4\ln{(x)}}{x} } \end{align*}

This is now in a form where L'Hospital's Rule can be applied.