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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?

Michael Hardy
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    Do you mean this? $$\lim_{x \to 1} \frac{\int_1^x \sin t dt}{x^2 - 1}$$ –  Apr 12 '14 at 22:15
  • Yes..I don't know how to type it that way sorry –  Apr 12 '14 at 22:15
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    You can definitely use l'Hôpital's theorem and the fundamental theorem of calculus; your computation seems good. – egreg Apr 12 '14 at 22:17

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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

$$\frac 1 {1 + 1} \cdot \frac{d}{dx} \left(\int_1^x \sin t dt\right)\Big|_{x = 1} = \frac 1 2 \sin 1$$