Limit Problem - # 7

[Jason76]

\(\displaystyle \lim x \rightarrow \infty\)

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

\(\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}}\) This seems to come out to \(\displaystyle 1\) as did the limit of (as x approached infinity) \(\displaystyle x^{\dfrac{4}{x}}\) But the computer says the answer is NOT 1. :-?

 

[Jason76]

Here is an answer I got on another forum:

\(\displaystyle \begin{align*} \lim_{x \to \infty} \left(e^x + x \right)^{9/x} & = \lim_{x \to \infty} \exp\left( \dfrac{9}{x}\ln(e^x + x)\right) \\ & = \exp\left( \lim_{x \to \infty} \dfrac{9\ln(e^x+x)}{x} \right) \\ & = \exp\left(\lim_{x \to \infty} \dfrac{ \tfrac{d}{dx}\left( 9\ln(e^x+x)\right) }{\tfrac{d}{dx}\left(x\right)}\right) \\ & = \exp\left(\lim_{x \to \infty} \dfrac{ \left(\dfrac{ 9(e^x + 1)}{e^x + x} \right) }{1} \right) \\ & = e^9\end{align*}\)

It's correct, but what is EXP? What is \(\displaystyle e^{\infty}\) This is all Greek to me. Any insight?

 

[fcabanski]

\(\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}}\) is \(\displaystyle ({\infty} + \infty)^{\dfrac{9}{\infty}}\) = \(\displaystyle ({\infty})^0\) which is an indeterminate form.

Use lim x--->inf (e^x + x)^(9/x) = e ^ (lim x ---> inf log((e^x + x) ^(9/x) ).

Let's remember that it's e raised to all that stuff, and just focus on the limit e is raised to.

lim x---> inf log((e^x + x)^(9/x)) = lim x--->inf 9log(e^x + x)/x

Now you can use L'Hospital's rule to find the limit.

 

[Jason76]

\(\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}}\) is \(\displaystyle ({\infty} + \infty)^{\dfrac{9}{\infty}}\) = \(\displaystyle ({\infty})^0\) which is an indeterminate form.

Use lim x--->inf (e^x + x)^(9/x) = e ^ (lim x ---> inf log((e^x + x) ^(9/x) ).

Let's remember that it's e raised to all that stuff, and just focus on the limit e is raised to.

lim x---> inf log((e^x + x)^(9/x)) = lim x--->inf 9log(e^x + x)/x

Now you can use L'Hospital's rule to find the limit.

But isn't \(\displaystyle \infty^{\dfrac{4}{\infty}} = \infty^{0} = 1\)? What is the difference between that problem and this one?

http://www.freemathhelp.com/forum/threads/84405-Limit-Problem-2

 

[Deleted member 4993]

But isn't \(\displaystyle \infty^{\dfrac{4}{\infty}} = \infty^{0} = 1\)? What is the difference between that problem and this one?

http://www.freemathhelp.com/forum/threads/84405-Limit-Problem-2

see

http://www.freemathhelp.com/forum/threads/84493-Confusing-Answers