TITLE: A limits question
QUESTION [7 upvotes]: The following equals?
$$
\lim_{x \to 1}\frac{\displaystyle\int_1^x \sin(t) \, dt}{x^2-1}
$$
I think this can be converted to
$$
\lim_{x \to 1}\frac{\sin(x)}{2x} = \frac{\sin(1)}{2}
$$
by using the fundamental theorem of calculus.
But the correct answer is $1/2$.
So where I made a mistake?

REPLY [3 votes]: What you've computed (apparently with L'Hospital's rule) is correct. For an alternative way, use the definition of the derivative:
we start by factoring the denominator to compute
$$\lim_{x \to 1} \frac{1}{x + 1} \cdot \frac{\int_1^x \sin t}{x - 1}$$
Computing each limit separately, the limit is equal to
$$\frac 1 {1 + 1} \cdot \frac{d}{dx} \left(\int_1^x \sin t dt\right)\Big|_{x = 1} = \frac 1 2 \sin 1$$