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DLS
 one year ago
Limits question
DLS
 one year ago
Limits question

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DLS
 one year 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
 one year 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!)

DLS
 one year ago
Best ResponseYou've already chosen the best response.1I THOUGHT U CRACKED IT >.<

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

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

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

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

DLS
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1I said simple because of my own conscience :P

terenzreignz
 one year 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
 one year 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
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1lengthy \\m// my method was better! :O

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

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

DLS
 one year 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
 one year 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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