Question

Limits question

Answer 1

\[\LARGE \lim_{x \rightarrow 0} (\frac{e^x+e^{-x}-2}{x^2})^\frac{1}{x^2}\]

Answer 2

@Callisto @satellite73

Answer 3

Summon @ParthKohli

Answer 4

(Just that you don't get in hopes: I'm just trying out. @Callisto please feel free to post your answer!)

Answer 5

I THOUGHT U CRACKED IT >.<

Answer 6

Questions from you guys are nightmare for me!

Answer 7

Summon @terenzreignz

Answer 8

Agree with @Callisto but let's see how this plays out :)

Answer 9

\[\LARGE e^{ (\frac{e^x+e^{-x}-2-x^2}{x^2})\frac{1}{x^2}}\]
this should be it,I just need to find the limit of this thing,hmm..

Answer 10

the power gives 1/12
hence e^(1/12) is the answer.

Answer 11

Isn't this pretty straightforward as far as L'Hôpital stuff is concerned?

Answer 12

\[\large { (\frac{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+-2-x^2}{x^2})\frac{1}{x^2}}\]
=112

Answer 13

i mean 1/12*

Answer 14

sorry i don't know why i posted such a simple ques :P maybe m sleepy .-.

Answer 15

just stuck me now

Answer 16

I didn't say simple, I said straightforward :D
\[\Large 1^\infty\]
So... suppose...
\[\Large L = \lim_{x\rightarrow 0}\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\]

Answer 17

I said simple because of my own conscience :P

Answer 18

Then...
\[\Large \ln(L) = \ln \lim_{x\rightarrow 0}\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\]
Since Logarithms are continuous, they may enter the limit...
\[\Large \ln(L) = \lim_{x\rightarrow 0} \ \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\]
\[\Large \ln(L) = \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)\]

Answer 19

\[\Large \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) =\lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}\]

And now we have an indeterminate form \(\large\frac00\)

Answer 20

LH twice? D:

Answer 21

hmm possibleway

Answer 22

Yup :)

Differentiating the numerator and denominator... (bloody hell, this'll be a mouthful :D )
\[\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]

Answer 23

\[\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]

Answer 24

lol

Answer 25

It keeps getting cropped >.>
\[\large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]

Answer 26

lengthy \\m// my method was better! :O

Answer 27

What was your method? I didn't catch it :D

Answer 28

expansions..

Answer 29

Series?

Answer 30

yes

Answer 31

Count me out of that one :P

Answer 32

hmm :P it was a single line solution with expansions,LH is deadly dangerous here.

Answer 33

That is true... unfortunately, I'm not well-versed with series...
I don't like series, series don't like me :D