## DLS 2 years ago Limits question

1. DLS

$\LARGE \lim_{x \rightarrow 0} (\frac{e^x+e^{-x}-2}{x^2})^\frac{1}{x^2}$

2. DLS

@Callisto @satellite73

3. DLS

Summon @ParthKohli

4. ParthKohli

(Just that you don't get in hopes: I'm just trying out. @Callisto please feel free to post your answer!)

5. DLS

I THOUGHT U CRACKED IT >.<

6. Callisto

Questions from you guys are nightmare for me!

7. DLS

Summon @terenzreignz

8. terenzreignz

Agree with @Callisto but let's see how this plays out :)

9. 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..

10. DLS

the power gives 1/12 hence e^(1/12) is the answer.

11. terenzreignz

Isn't this pretty straightforward as far as L'Hôpital stuff is concerned?

12. 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

13. DLS

i mean 1/12*

14. DLS

sorry i don't know why i posted such a simple ques :P maybe m sleepy .-.

15. DLS

just stuck me now

16. terenzreignz

I didn't say simple, I said straightforward :D $\Large 1^\infty$ So... suppose... $\Large L = \lim_{x\rightarrow 0}\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}$

17. DLS

I said simple because of my own conscience :P

18. terenzreignz

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... $\Large \ln(L) = \lim_{x\rightarrow 0} \ \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}$ $\Large \ln(L) = \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)$

19. terenzreignz

$\Large \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) =\lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}$ And now we have an indeterminate form $$\large\frac00$$

20. DLS

LH twice? D:

21. DLS

hmm possibleway

22. terenzreignz

Yup :) Differentiating the numerator and denominator... (bloody hell, this'll be a mouthful :D ) $\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}$

23. terenzreignz

$\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}$

24. DLS

lol

25. terenzreignz

It keeps getting cropped >.> $\large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}$

26. DLS

lengthy \\m// my method was better! :O

27. terenzreignz

What was your method? I didn't catch it :D

28. DLS

expansions..

29. terenzreignz

Series?

30. DLS

yes

31. terenzreignz

Count me out of that one :P

32. DLS

hmm :P it was a single line solution with expansions,LH is deadly dangerous here.

33. terenzreignz

That is true... unfortunately, I'm not well-versed with series... I don't like series, series don't like me :D