Question

Immediate. Immediate. Please help with limit problem?

Answer 1

`lim_(x->0)(e^x-e^-x)/x=2`

Explanation:

We want to solve

`lim_(x->0)(e^x-e^-x)/x`

This leads to an indeterminate form `0/0`,
therefore we can apply L'Hôpital's rule

`color(blue)(lim_(x->a)f(x)/g(x)=lim_(x->a)(f'(x))/(g'(x))`

Thus

`lim_(x->0)(e^x-e^-x)/x=lim_(x->0)(e^x+e^-x)/1=2`

Answer 2

`2`

Explanation:

`e^x = sum_(k=0)^oo x^k/(k!)`
and
`e^-x = sum_(k=0)^oo (-x)^k/(k!)`

then `e^x-e^-x = 2sum_(k=0)^oo x^(2k+1)/((2k+1)!) =2x(1+f(x))`

then

`lim_(x->0)(e^x-e^-x)/x = lim_(x->0)2(x(1+f(x)))/x = lim_(x->0) 2(1+f(x))=2`

Answer 3

Image....

Explanation:

Without L Hospital rule...