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  • DLS

Limits question

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

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Other answers:

(Just that you don't get in hopes: I'm just trying out. @Callisto please feel free to post your answer!)
  • DLS
I THOUGHT U CRACKED IT >.<
Questions from you guys are nightmare for me!
  • DLS
Agree with @Callisto but let's see how this plays out :)
  • DLS
\[\LARGE e^{ (\frac{e^x+e^{-x}-2-x^2}{x^2})\frac{1}{x^2}}\] this should be it,I just need to find the limit of this thing,hmm..
  • DLS
the power gives 1/12 hence e^(1/12) is the answer.
Isn't this pretty straightforward as far as L'Hôpital stuff is concerned?
  • DLS
\[\large { (\frac{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+-2-x^2}{x^2})\frac{1}{x^2}}\] =112
  • DLS
i mean 1/12*
  • DLS
sorry i don't know why i posted such a simple ques :P maybe m sleepy .-.
  • DLS
just stuck me now
I didn't say simple, I said straightforward :D \[\Large 1^\infty\] So... suppose... \[\Large L = \lim_{x\rightarrow 0}\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\]
  • DLS
I said simple because of my own conscience :P
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... \[\Large \ln(L) = \lim_{x\rightarrow 0} \ \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\] \[\Large \ln(L) = \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)\]
\[\Large \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) =\lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}\] And now we have an indeterminate form \(\large\frac00\)
  • DLS
LH twice? D:
  • DLS
hmm possibleway
Yup :) Differentiating the numerator and denominator... (bloody hell, this'll be a mouthful :D ) \[\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]
\[\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]
  • DLS
lol
It keeps getting cropped >.> \[\large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]
  • DLS
lengthy \\m// my method was better! :O
What was your method? I didn't catch it :D
  • DLS
expansions..
Series?
  • DLS
yes
Count me out of that one :P
  • DLS
hmm :P it was a single line solution with expansions,LH is deadly dangerous here.
That is true... unfortunately, I'm not well-versed with series... I don't like series, series don't like me :D

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