# Math Help - e Limit Problem

1. ## e Limit Problem

$\lim x \rightarrow \infty$

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

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

2. ## Re: e Limit Problem

\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*}

3. ## Re: e Limit Problem

But how come it didn't evaluate to 1 from the get go? Why did we proceed to the derivative and then plugin?

What is $e^{\infty}$ ?

4. ## Re: e Limit Problem

Slip eternal,
You and I must have been working on this problem at the same time. I did it differently

lim [ex+x]9/x
x->infinity

=lim ex(1+(x/ex )]9/x
x->infinity

=lim ex9/x(1+(x/ex )]9/x
x->infinity

=lim e9(1+(x/ex )]9/x
x->infinity

= e9*1

= e9

Extra Little Note:
SlipEternal turned it into a algebraic fraction so she/he could use L'hopital's Rule to solve it.
L'Hôpital's rule - Wikipedia, the free encyclopedia

5. ## Re: e Limit Problem

Originally Posted by Jason76
But how come it didn't evaluate to 1 from the get go? Why did we proceed to the derivative and then plugin?

What is $e^{\infty}$ ?
It didn't evaluate to 1 from the get go because you cannot plug in infinity. Infinity is not a number. $\exp(f(x))$ is another way of writing $e^{f(x)}$. I used that notation because I wanted to be clear what was in the exponent of e. So, I did $e^{\ln(\text{original expression})}$. That was how I started. Then, using the properties of natural log, the exponent becomes a coefficient. Since $e^{f(x)}$ is a composition of continuous functions, I used the limit law that states that if $g$ is a continuous function, then $\lim_{x \to a} g(f(x)) = g\left(\lim_{x \to a}f(x)\right)$. That's how I brought the limit inside the $\exp(\cdots)$ function. Then, on line 2, the expression $\dfrac{9\ln(e^x + x)}{x}$ is approaching $\dfrac{\infty}{\infty}$ as $x \to \infty$, so we can apply L'Hospital's Rule. I took the derivative of the top and the derivative of the bottom. Then, I skipped a step. We had $\dfrac{9(e^x+1)}{e^x+x} = \dfrac{9+\tfrac{1}{e^x}}{1+\tfrac{x}{e^x}}$. As $x \to \infty$, $\tfrac{1}{e^x}\to 0$ and $\tfrac{x}{e^x} \to 0$. So, $\exp\left(\lim_{x\to \infty} \dfrac{9+\tfrac{1}{e^x}}{1+\tfrac{x}{e^x}} \right) = \exp(9) = e^9$.

To recap, $e^\infty$ is not defined because $\infty$ is not a number. So, instead, I altered the form so that I could apply L'Hospital's Rule.

6. ## Re: e Limit Problem

lim [ex+x]9/x
x->infinity

I have a slightly different explanation of why you can't just substitute in infinity.

ex+x is infinitely large as x approaches infinity.

9/x is infinitely small as x approaches infinity.

They are not approaching infinity or 0 at the same rate. You have to change the form or it will not be valid.

Also,

Using your logic I think you might be tempted to say that

lim [ex]9/x = 1
x->infinity

but you can see that it can be transformed into

lim e9x/x
x->infinity

lim e9
x->infinity

= e9

Think it over, I hope that it will help.

7. ## Re: e Limit Problem

Originally Posted by Melody2
Slip eternal,
You and I must have been working on this problem at the same time. I did it differently

lim [ex+x]9/x
x->infinity

=lim ex(1+(x/ex )]9/x
x->infinity

=lim ex9/x(1+(x/ex )]9/x
x->infinity

=lim e9(1+(x/ex )]9/x
x->infinity

= e9*1

= e9

Extra Little Note:
SlipEternal turned it into a algebraic fraction so she/he could use L'hopital's Rule to solve it.
L'Hôpital's rule - Wikipedia, the free encyclopedia
To clear up the notation (matching brackets and parentheses), here is Melody2's solution again (which does not need L'Hospital's Rule):

\begin{align*}\lim_{x \to \infty}(e^x+x)^{9/x} & = \lim_{x \to \infty}\left[e^x\left(1+\dfrac{x}{e^x}\right) \right]^{9/x} \\ & = \lim_{x \to \infty}e^{9x/x}\left(1+\dfrac{x}{e^x}\right)^{9/x} \\ & = \lim_{x \to \infty} e^9\left(1+\dfrac{x}{e^x}\right)^{9/x} \\ & = e^9 \cdot 1 \\ & = e^9\end{align*}

8. ## Re: e Limit Problem

Thanks SlipEternal,
how do you write it out neatly like that?

9. ## Re: e Limit Problem

Originally Posted by Melody2
Thanks SlipEternal,
how do you write it out neatly like that?
Hit "Reply With Quote" on my previous post to see what I entered. I am using LaTeX formatting.

10. ## Re: e Limit Problem

I have just down loaded Latex but I haven't tried to use it yet. I have questions but I probably need to play with it first.
Thanks