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DLSBest ResponseYou've already chosen the best response.1
\[\LARGE \lim_{x \rightarrow 0} (\frac{e^x+e^{x}2}{x^2})^\frac{1}{x^2}\]
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
(Just that you don't get in hopes: I'm just trying out. @Callisto please feel free to post your answer!)
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
I THOUGHT U CRACKED IT >.<
 10 months ago

CallistoBest ResponseYou've already chosen the best response.0
Questions from you guys are nightmare for me!
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Agree with @Callisto but let's see how this plays out :)
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\LARGE e^{ (\frac{e^x+e^{x}2x^2}{x^2})\frac{1}{x^2}}\] this should be it,I just need to find the limit of this thing,hmm..
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
the power gives 1/12 hence e^(1/12) is the answer.
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Isn't this pretty straightforward as far as L'Hôpital stuff is concerned?
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
\[\large { (\frac{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+1x+\frac{x^2}{2}\frac{x^3}{6}+\frac{x^4}{24}+2x^2}{x^2})\frac{1}{x^2}}\] =112
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
sorry i don't know why i posted such a simple ques :P maybe m sleepy ..
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
I didn't say simple, I said straightforward :D \[\Large 1^\infty\] So... suppose... \[\Large L = \lim_{x\rightarrow 0}\left(\frac{e^x +e^{x}2}{x^2}\right)^{\frac1{x^2}}\]
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
I said simple because of my own conscience :P
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
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... \[\Large \ln(L) = \lim_{x\rightarrow 0} \ \ln\left(\frac{e^x +e^{x}2}{x^2}\right)^{\frac1{x^2}}\] \[\Large \ln(L) = \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{x}2}{x^2}\right)\]
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
\[\Large \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{x}2}{x^2}\right) =\lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{x}2}{x^2}\right) }{x^2}\] And now we have an indeterminate form \(\large\frac00\)
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Yup :) Differentiating the numerator and denominator... (bloody hell, this'll be a mouthful :D ) \[\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{x}2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{x}2}\right)\left(\frac{x^2(e^xe^{x})2x(e^x+e^{x}2)}{x^4}\right)}{2x}\]
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
\[\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{x}2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{x}2}\right)\left(\frac{x^2(e^xe^{x})2x(e^x+e^{x}2)}{x^4}\right)}{2x}\]
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
It keeps getting cropped >.> \[\large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{x}2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{x}2}\right)\left(\frac{x^2(e^xe^{x})2x(e^x+e^{x}2)}{x^4}\right)}{2x}\]
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
lengthy \\m// my method was better! :O
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
What was your method? I didn't catch it :D
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Count me out of that one :P
 10 months ago

DLSBest ResponseYou've already chosen the best response.1
hmm :P it was a single line solution with expansions,LH is deadly dangerous here.
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
That is true... unfortunately, I'm not wellversed with series... I don't like series, series don't like me :D
 10 months ago
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