Question

limit

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \sin(e^{x^2}-1) }{ x^2 }\]

Answer 2

try out l hospitals rule

Answer 3

@Jonask

Answer 4

you can find answers in algebra books..

Answer 5

multiply divide by e^(x^2) -1 to directly reach the ans..

Answer 6

as we know these things :

1) e^(x^2) -1 --> 0 ,for x-->0
2)e^x - 1 /x = 1 for x-->0
3) sin x / x = 1 for x-->0

hope that;d help..

Answer 7

simply apply L HOSPITAL rule.

Answer 8

it says without the rule,i did this
\[\frac{\sin(x^2(\frac{ e^{x^2}-1 }{ x^2 }))}{x^2}\]
since\[\lim_{h \rightarrow 0} \frac{e^h-1}{h}=1\]\[\frac{\sin (x^2(1))}{x^2}=1\]

Answer 9

same result

Answer 10

as what