MIT 18.02 Multivariable Calculus, Fall 2007
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (jamesj):

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.

OpenStudy (jamesj):

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

OpenStudy (jamesj):

the limit of those expression as x --> pi

OpenStudy (anonymous):

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

OpenStudy (anonymous):

0, but denominator cant be 0 right?

OpenStudy (jamesj):

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

OpenStudy (anonymous):

cosx*e^(sinx)

OpenStudy (jamesj):

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

OpenStudy (anonymous):

1

OpenStudy (jamesj):

-1

OpenStudy (anonymous):

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.

OpenStudy (jamesj):

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.

OpenStudy (jamesj):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok thx