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DLS
 3 years ago
Limits question
DLS
 3 years ago
Limits question

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

ParthKohli
 3 years ago
Best 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!)

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0Questions from you guys are nightmare for me!

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Agree with @Callisto but let's see how this plays out :)

DLS
 3 years ago
Best 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..

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1the power gives 1/12 hence e^(1/12) is the answer.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Isn't this pretty straightforward as far as L'Hôpital stuff is concerned?

DLS
 3 years ago
Best 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

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1sorry i don't know why i posted such a simple ques :P maybe m sleepy ..

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1I 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
 3 years ago
Best ResponseYou've already chosen the best response.1I said simple because of my own conscience :P

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Then... \[\Large \ln(L) = \ln \lim_{x\rightarrow 0}\left(\frac{e^x +e^{x}2}{x^2}\right)^{\frac1{x^2}}\] Since Logarithms are continuous, they may enter the limit... \[\Large \ln(L) = \lim_{x\rightarrow 0} \ \ln\left(\frac{e^x +e^{x}2}{x^2}\right)^{\frac1{x^2}}\] \[\Large \ln(L) = \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{x}2}{x^2}\right)\]

terenzreignz
 3 years ago
Best 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 \(\large\frac00\)

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

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

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

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1lengthy \\m// my method was better! :O

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1What was your method? I didn't catch it :D

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Count me out of that one :P

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

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