I was doing Indeterminate Form. I saw something weird while differentiating using product rule.
$$f(x)g(x)=f'(x)g(x)+f(x)g'(x)$$
I got a value which was looking like.
$$\lim_{x->0} \frac{e^x-e^{\sin x }\cos^2x}{\sin x} +\lim_{x->0} e^{\sin x}$$
So, I have to differentiate up and down cause, it is indeterminate form. [first limit is $\frac{0}{0}$] $$=\lim_{x->0} \frac{e^x-(e^{\sin x }\cos^3x+2e^{\sin x} \cos x \sin x)}{\cos x} +\lim_{x->0} e^{\sin x}$$
When I used product rule on numerator the value was looking like this. But, my book wrote that there will a negative before first one there won't negative before second one.
Your mistake is from $$\lim_{x->0} \frac{e^x-e^{\sin x }\cos^2x}{\sin x} +\lim_{x->0} e^{\sin x}$$
We know that derivative of $\cos x$ is $-\sin x$
So it would be
$$=\lim_{x->0} \frac{e^x-(e^{\sin x }\cos^3x-2e^{\sin x} \cos x \sin x)}{\cos x} +\lim_{x->0} e^{\sin x}$$
Your mistake was you took derivative of $\cos x$ as $\sin x$