Limits 2

This is wrong The derivative of sin(hx) is hcos (hx) so limit will be h

The limit of $\displaystyle\frac {sinhx}{x}$ as x approachs to 0 will result in $\frac{0}{0}$ (an indeterminate form). So, we need to derivate the top and bottom and take limit.

The derivative of $\displaystyle\sinh x = \displaystyle cosh 0 = 1$ , and the derivative of $0 = 1$ . So, the limit is $\frac {1}{1} = \boxed {1}$

sinh x=(e^x-e^(-x))/2