Math Help - L'hopital question?

1. L'hopital question?

Lim (e^(x)+x)^(1/x)
x->∞

What I came to was e^(1). I'm not sure if this is the right answer, the TI-83 has it at e^(1) up to 230, then at 231 it's undefined. I don't understand why, my TI-89 has it as ∞ past 231.

2. Originally Posted by Myung
Lim (e^(x)+x)^(1/x)
x->∞

What I came to was e^(1). I'm not sure if this is the right answer, the TI-83 has it at e^(1) up to 230, then at 231 it's undefined. I don't understand why, my TI-89 has it as ∞ past 231.

but

$\lim_{x\to\infty}\left(e^x+x\right)^{\frac{1}{x}}= \lim_{x\to\infty}\frac{e^{x+1}+x+1}{e^x+x}=e$

So I solved it in a non-L'hopital's manner, so you show us your way, and we will help with snags.

3. Sorry about the formatting, I couldn't get the formatting to work, tried copy pasting math codes but it came out in error. Anyway, here goes:

e^ln(e^x+x)^(1/x) = e^(e^x+1)/(e^x+x) = e^(e^x/(e^x+1)) = e^(1), as x approaches infinity. The 83 gives undefined as x is greater than 230 and 89 has it as infinity, but at 230 it's e^(1). It doesn't make sense to me.

4. Originally Posted by Myung
Sorry about the formatting, I couldn't get the formatting to work, tried copy pasting math codes but it came out in error. Anyway, here goes:

e^ln(e^x+x)^(1/x) = e^(e^x+1)/(e^x+x) = e^(e^x/(e^x+1)) = e^(1), as x approaches infinity. The 83 gives undefined as x is greater than 230 and 89 has it as infinity, but at 230 it's e^(1). It doesn't make sense to me.
Hmm, I am not sure what you did between your first step and your second step,

try from this point

$L=\lim_{x\to\infty}\left(e^x+x\right)^{\frac{1}{x} }\Rightarrow\ln(L)=\lim_{x\to\infty}\frac{\ln\left (e^x+x\right)}{x}$

5. I found the derivative of lnx and x.

e^x+1 / e^x+x then did it again, e^x / e^x+1

then set the whole thing to e = e^(1)

6. Originally Posted by Myung
I found the derivative of lnx and x.

e^x+1 / e^x+x then did it again, e^x / e^x+1

then set the whole thing to e = e^(1)
Ok great! My bad, I didnt see what you were doing, so as you noticed

$\lim_{x\to\infty}\frac{\ln\left(e^x+x\right)}{x}=\ lim_{x\to\infty}\frac{\frac{e^x+1}{e^x+x}}{1}=1$

And also as you stated, remebering that $L$ is our orginal limit if

$\ln(L)=1\Rightarrow{L=e^1=e}$

7. Do you know why the calculator has it as undefined at anything past 230 or why the 89 has infinity as the answer past 230? I don't see any critical points past 230 or any discrepencies.

8. Originally Posted by Myung
Do you know why the calculator has it as undefined at anything past 230 or why the 89 has infinity as the answer past 230? I don't see any critical points past 230 or any discrepencies.
Ok, let $f(x)=(e^x+x)^{\frac{1}{x}}$

My 89 gives $f(230)=2.71828$
$f(2300)=2.71828$

$f(2300000)=2.71828$

but

$f(23000000)=\infty$

I assume that the caclulator does not have enough memory to do this, so it assumes it is zero.

9. Well that is very odd. I was completely sure e^(1) was the answer algebraically, I just couldn't figure out why the calculator was trying to throw me off. I use the calculator as a "check" to make sure I have the right answer, but this case it didn't give me the right answer. Still a bit puzzled as to why this problem will do that, since as x-> infinity, y is approaching a constant value eh? Maybe it's the e that is throwing it off? Most other problems tend to give the right answer for horizontal limits.

10. Originally Posted by Myung
Well that is very odd. I was completely sure e^(1) was the answer algebraically, I just couldn't figure out why the calculator was trying to throw me off. I use the calculator as a "check" to make sure I have the right answer, but this case it didn't give me the right answer. Still a bit puzzled as to why this problem will do that, since as x-> infinity, y is approaching a constant value eh? Maybe it's the e that is throwing it off? Most other problems tend to give the right answer for horizontal limits.
Also be weary for, some limits converge very slowly

So $f\left(10^{10^{10}}\right)$ might not be close to $\lim_{x\to\infty}f(x)$

Also note that this will method of plug and chug will give you an approximation, so the answer might be $\sqrt{2}\pi$ but you would never guess that from your evaluation giving you $4.4428$

11. What I do is, if I do it algebraically and get sqrt2 pi, I will plug the equation into my calculator and see what it gives me. If it gives me 4.4428, I will approximate sqrt2 pi into my calculator to see if it gives me 4.4428, if it does, then I may have the right answer. Is there a reason I would want to use 10^10^10 as x? I usually just use 999 or anything random.

12. Originally Posted by Myung
What I do is, if I do it algebraically and get sqrt2 pi, I will plug the equation into my calculator and see what it gives me. If it gives me 4.4428, I will approximate sqrt2 pi into my calculator to see if it gives me 4.4428, if it does, then I may have the right answer. Is there a reason I would want to use 10^10^10 as x? I usually just use 999 or anything random.
Yeah,

Let $x>y$

Then obviously $f(x)$ is going to be a better approximation than $f(y)$ because $x$ is closer to $\infty$ than $y$

13. Ok that is good to know.