Question

evaluate limit

Answer 1

Since both top and bottom evaluate to \(0\), we can say \(\lim_{x\to0}f(x)=\lim_{x\to0}g(x)=0\) and \(f(x)=e^x+e^{-x}-2,g(x)=x^2\). This qualifies for l'Hopital's rule, so:\[\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}=\lim_{x\to0}\frac{e^x-e^{-x}}{2x}=\lim_{x\to0}\frac{e^x+e^{-x}}{2}\]the last part being aparent from reapplying l'Hopital's.

Answer 2

how to see it in a better way its hectic

Answer 3

It's a bit complicated, yes, because we're applying the same rule twice. I'm fairly certain that this is the simplest solution, unfortunately.

Just remember the second application of l'Hopital's qualifies because \(\lim_{x\to0}f'(x)=\lim_{x\to0}g'(x)=0\) as well.

Answer 4

If you're having a bit of trouble understanding a certain part, ask :D

Answer 5

i am not able to read your answer its too complicated, and unclear...

Answer 6

can u do it in a easy-to -read way

Answer 7

Okay, then let me write this step by step. l'Hopital's rule states that if\[\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0,\infty\]then\[\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\]Now, here our \(f(x)=e^x+e^{-x}-2\) and \(g(x)=x^2\); most notably, they both converge to \(0\) at \(x\to0\). Since \(\frac{0}{0}\) is undefined, let's try applying l'Hopital's rule.\[\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{e^x+e^{-x}-2}{x^2}=\lim_{x\to0}\frac{e^x-e^{-x}}{2x}\]Since top and bottom still converge to \(0\) as \(x\to0\), we can extend l'Hopital's:\[\lim_{x\to a}f'(x)=\lim_{x\to a}g'(x)=0,\infty\]thus\[\lim_{x\to a}\frac{f'(x)}{g'(x)}=\lim_{x\to a}\frac{f''(x)}{g''(x)}\]and therefore, repeating the application:\[\lim_{x\to0}\frac{e^x+e^{-x}-2}{x^2}=\lim_{x\to0}\frac{e^x-e^{-x}}{2x}=\lim_{x\to0}\frac{e^x+e^{-x}}{2}\]so now it should be trivial to solve. :D

Answer 8

Do you understand?

Answer 9

If you don't, I'm sorry. I need to leave do my own work. XD Good luck with this!