Is it possible to solve this limit without using L'hopital?

prob has to do with using hyperbolic trig identities on the denominator i don't fuggin know

http://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+of+x+sin+%282x%29+%2F+%28e%5Ex+%2B+e%5E-2+%2B2%29

>>7686637
0sin(2(0))/(1+1-2)=0

>>7686653
Oh i see where its fucking up.

Can you take the conjugate of the denominator?

At least you know it approaches 2.

But now you gotta do the work

File: Screenshot_2015-11-28-13-40-48.png (413KB, 1440x2560px) Image search: [Google]

413KB, 1440x2560px

>>7686665

>>7686667
what's the app?

is it note 5 btw?

>>7686653
Indetermination

>>7686665
Yeah, but I don't even have a clue of how to operate those "e".

damn, for this ecuation I lost 1 year of university.

>>7686642
Dude that was so funny.

>>7686637
Multiply the numerator by 2x/2x, you will have 2x^2 * [sin(2x)/2x]/(e^x + e^-x -2). Using the multiplication theorem for limits of continuous functions, we have it that this limit is the same as [limit as x approaches 0 of 2x^2/(e^x + e^-x -2) times the limit as x approaches 0 of sin(2x)/2x. Using epsilon-delta, the right limit is the same as sin(y)/y as y approaches 0, so we have limit as x approaches 0 of 2x^2/(e^x + e^-x - 2) * 1. Using taylor series, e^x + e^-x -2 = 2*x^2/2! + 2*x^4/4! + 2*x^6/6!....Dividing the numerator and denominator by 2x^2, we get 1/2 + x^2/4! + x^4/6! ....This series is bounded from above by a geometric series (1/2 + (x^n - x^2)/(x^2 - 1)), and bounded from below by 1/2, so, taking the limit as x approaches 0, the series is bounded from above by 1/2 + (0-0)/(0-1), and bounded from below by 1/2, so the limit is 1/(1/2) = 2.

>>7686693
Actually, the geometric series is equal to x^2/(1-x^2) for |x|<1, the x^n term shouldn't be there.

I guess you have to do a Taylor approximation of the upper and lower function around x=0

>>7686673
It's the mathlab graphing calculator for Android. The free version is pretty good, the paid version gets 3D graphs

>>7686720
nice, or just plug x=0 into the (continuous) power series in the denominator to get 1/2 + 0 + 0 + \dotsb

>>7686637
>without using L'hopital
Find the first few terms of it's Taylor Series and then evaluate the limit.

>>7686693
A better argument for this is the Taylor series identity limit as x approaches a of R_n,a(x)/(x-a)^n = 0, where R_n,a is the Taylor series remainder for n terms, centered at a. With this, e^x and e^-x can be replaced with 1+x+x^2/2 + R(e^x)_2,0(x) and 1-x+x^2 + R(e^-x)_2,0(x), so, if we divide the numerator and denominator by 2x^2, we get limit as x approaches 0 of [sin(2x)/2x]/(1/4+1/4 +[R(e^x)_2,0(x)/(x-0)^2]/2 + [R(e^-x)_2,0(x)/(x-0)^2]/2). With our limit identity, and the sin(y)/y limit, we get 1/(1/4+1/4+0/2+0/2) = 2.