Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

DLS Group TitleBest ResponseYou've already chosen the best response.1
@Callisto @satellite73
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
Summon @ParthKohli
 one year ago

ParthKohli Group TitleBest 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!)
 one year ago

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

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

DLS Group TitleBest ResponseYou've already chosen the best response.1
Summon @terenzreignz
 one year ago

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

DLS Group TitleBest 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..
 one year ago

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

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

DLS Group TitleBest 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
 one year ago

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

DLS Group TitleBest ResponseYou've already chosen the best response.1
just stuck me now
 one year ago

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

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

terenzreignz Group TitleBest 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... \[\Large \ln(L) = \lim_{x\rightarrow 0} \ \ln\left(\frac{e^x +e^{x}2}{x^2}\right)^{\frac1{x^2}}\] \[\Large \ln(L) = \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{x}2}{x^2}\right)\]
 one year ago

terenzreignz Group TitleBest 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 \(\large\frac00\)
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.1
hmm possibleway
 one year ago

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

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

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

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

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Series?
 one year ago

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

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

terenzreignz Group TitleBest 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
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.