Evaluate the limit $lim_(x to pi) (e^sinx -1)/(x-pi)$ without using L'Hôpital's rule?

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Answer 1 · 2018-05-13T21:21:40

$-1$

$lim_(x to pi) ((e^sin(x)-1)/(x-pi))$

With $y = x - pi$

$= lim_(y to 0) ((e^sin(y + pi)-1)/(y))$

Aside:

$= lim_(y to 0) ((e^(-sin(y))-1)/(y))$

If we can use the definition of the exponential function, ie:

$e^z=\sum _{k=0}^{oo }z^{k}/(k!)=1+z+z^{2}/2 +z^{3}/6 +\cdots$

Then $e^(-sin(y))$ can be expanded as:

$= lim_(y to 0) ((1 -sin(y) + 1/2 (-sin^2 y) + .... )-1)/(y)$

$= lim_(y to 0) ( -sin(y) + 1/2 sin^2 y + .... )/(y)$

$=- lim_(y to 0) ( sin(y) - 1/2 sin^2 y + .... )/(y) = -1$

Evaluate the limit lim_(x to pi) (e^sinx -1)/(x-pi)lim_(x to pi) (e^sinx -1)/(x-pi) without using L'Hôpital's rule?