prove the following statement, lim x^2/(e^x^2-1) = 1 where x-->0?

Answer 1

Try using the L'Hospital rule, by replacing the numerator and the denominator with their respective derivatives:
lim x^2/(e^x^2-1) = lim 2x/(2xe^x^2) = lim 1/e^x^2 = 1

Answer 2

Another solution, besides L'Hopital, arises from the fact that e^x = 1 + x + o(x), where o is a function such that o(x)/x --> 0 as x --> 0. So, (x)/(e^x -1) =(x)/(x + o(x)) = 1/((1 + o(x)/x), so that lim (x --> 0) (x)/(e^x -1) = 1(1 + 0) = 1.

Since x^2 vanishes at 0 and is continuous, and (y)/(e^y -1) --> 1 as y --> 0, it follows lim (x^2)/(e^(x^2)-1) = 1

Answer 3

bleedin hell! This is how to cheat at math
e^x^2-1=x^2+(x^2)^2+(x^2)^3/3!+..higher terms in x^2n/n!..( think its a Mclaur exp or summat)
So..and here's the dodgy bit..obv the higher terms in x disappear more quickly as x approaches 0,(negligible I think it's called) so u r left with lim= x^2/x^2 = 1!!

Answer 4

the limit goes to 0/0 as x goes to zero. So apply L'Hoptitals theorem.

take derivatives of the top and bottom (separately...)

2*x/(2xe^x^2) (I am assuming e^x^2 = e^(x^2) and not (e^x)^2)

cancel out the 2x terms in top and bottom

1/e^x^2 which goes to 1 as x goes to zero.