1
$\begingroup$

I was reviewing some of my notes from Calculus 1 so that I can prepare for Calculus 2 this fall, and I ran into one problem where I don't understand how the factoring works.

$$\lim_{x\to\infty} \frac{e^{1/x}+e^{-1/x}}{e^{1/x}-e^{-1/x}}$$

Then substituting... $$u = \frac {1}{x}$$

I know what the answer to this question is, but this step in the solution manual confused me:

$$\lim_{x\to\infty} \frac{e^u(1+e^{-2u})}{e^u(1-e^{-2u})}$$

My brain is still getting prepared to get back into doing math regularly after the summer vacation, but I'm hitting a serious wall trying to understand how this expression can be factored this way.

Any help in understanding this is greatly appreciated!

$\endgroup$
  • 2
    $\begingroup$ First there isn't any $n$ in the expression after the lim symbol. But it also looks like you miscopied something from the solution manual. I think the top factor should be $(1+e^{-2u})$. That way when "multiplied out" the top becomes $e^u+e^{-u}$ as it should to match. $\endgroup$ – coffeemath Aug 10 '13 at 3:15
  • $\begingroup$ @coffeemath good catch! Both typos. I will fix now $\endgroup$ – hax0r_n_code Aug 10 '13 at 3:17
2
$\begingroup$

HINT: If $u=1/x$ then $e^u(1+e^{-2u})=e^u+e^{-u}=e^\frac1x+e^{-\frac1x}$ which now matches the top of your expression. Similarly $e^u(1-e^{2u})$ matches the bottom of the expression. In the $u$ version the two $e^u$ factors cancel, and the expression becomes $$\frac{1+e^{-2/x}}{1-e^{-2/x}}.$$ This looks like, as $x \to \infty$, then $-2/x \to 0$ so that the exponential terms go to 1 and it diverges.

$\endgroup$
  • 2
    $\begingroup$ Actually, it does not since $\lim_{x \to \infty} (1+e^{-\frac{2}{x}}) = 2 \neq 0$ $\endgroup$ – Doctor Dan Aug 10 '13 at 3:28
  • $\begingroup$ @DoctorDan Correct comment. So it diverges... $\endgroup$ – coffeemath Aug 10 '13 at 3:31
1
$\begingroup$

if you have substituted u = 1/x then your limit is now u → 0 and e-2u → 1. The top term → 2 and the bottom term to -∞. So yes it diverges. Coffeemath said the same thing, but I want to point out to be careful that when you change variables you also must change the limits accordingly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.