# Thread: Limit Problem

1. ## Limit Problem

$\displaystyle \lim x \rightarrow \infty$

$\displaystyle (e^{x} + x)^{\dfrac{9}{x}}$

$\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}}$

$\displaystyle (\infty + \infty)^{0}$

$\displaystyle (\infty)^{0}$

2. ## Re: Limit Problem

Given $\displaystyle \lim_{x\rightarrow \infty}\left(e^x+x\right)^{\frac{9}{x}}$

Now when $\displaystyle x\rightarrow \infty$, Then $\displaystyle e^x>>>>>>>>>x$

So $\displaystyle \lim_{x\rightarrow \infty}\left(e^x+x\right)^{\frac{9}{x}}\approx \lim_{x\rightarrow \infty}\left(e^x\right)^{\frac{9}{x}} = e^9$

3. ## Re: Limit Problem

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

$\displaystyle (e^{x} + x)^{\dfrac{9}{x}}$

$\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}}$

$\displaystyle (\infty + \infty)^{0}$

$\displaystyle (\infty)^{0}$
For a more sophisticated approach

\displaystyle \displaystyle \begin{align*} \left( e^x + x \right) ^{\frac{9}{x}} &= e^{\ln{\left[ \left( e^x + x \right) ^{\frac{9}{x}} \right] }} \\ &= e^{\frac{9\ln{ \left( e^x + x \right) }}{x}} \end{align*}

and so

\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty} \left[ \left( e^x + x \right) ^{\frac{9}{x}} \right] &= \lim_{x \to \infty} e^{ \frac{9\ln{ \left( e^x + x \right) }}{x} } \\ &= e^{ \lim_{x \to \infty} \frac{9\ln{ \left( e^x + x \right) }}{x} } \\ &= e^{ \lim_{x \to \infty} \frac{\frac{9 \left( e^x + 1 \right) }{e^x + x}}{1} } \textrm{ by L'Hospital's Rule} \\ &= e^{ \lim_{ x \to \infty} \frac{9 \left( e^x + 1 \right) }{e^x + x} } \\ &= e^{\lim_{x \to \infty} \frac{9e^x}{e^x + 1} } \textrm{ by L'Hospital's Rule again} \\ &= e^{\lim_{x \to \infty} \frac{9e^x}{e^x} } \textrm{ by L'Hospital's Rule again} \\ &= e^{\lim_{x \to \infty} 9 } \\ &= e^9 \end{align*}