find the limx->pi (e^(sinx) -1)/(x-pi) @Mathematics

First look at the denominator. When x is approaching pi, and you're subtracting pi, what does the bottom go to?

By definition this is the derivative of f(x) = e^(sin x) at x = pi. Hence, instead of calculating the limit directly, write down the derivative of that function and evaluate it at x = pi.

That is: \[ f'(\pi) = \frac{ f(x) - f(\pi)}{x-\pi} = \frac{ e^{\sin x}- 1}{x-\pi} \]

the limit of those expression as x --> pi

^^Right, so when you take the limit as x-> pi on the denominator, what does it equal?

0, but denominator cant be 0 right?

@eigenvector, I don't know where you're going. This limit does exist. @veryconfused. What is the derivative of f(x) = e^(sin x)?

cosx*e^(sinx)

Right and value of f'(x) for x = pi?

1

-1

When you take the limit as x-> pi, and you're subtracting pi from x, the denominator goes to 0. You're making this too complicated. Just use L'Hopitals.

On the contrary, using l'Hopital's rule here relies on the derivative of exactly the same function and evaluating it at exactly the same value. So in fact, it's a circular logic to use l'Hopital's rule to evaluate the limit that defines a derivative to find that derivative.

\[ \lim_{x \rightarrow \pi} \frac{ f(x) - f(\pi)}{x - \pi} = f'(\pi) = -1\] Hence the answer is -1.

o yaaaa,, cos pi = -1, thus cosxe^sinx = -1 my bad

ok thx

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