Calculus problem #2201

Let $f(t)= \int_{3}^{t} \frac{4x-2}{x^{2}+1}\ dx$ . Then $f'(t)=\frac{4t-2}{t^{2}+1}$ and it is continuous at any real number.Then the limit in the question can be written like this $L=\lim_{h\to 0} \frac {f(3+h)-f(3-h)}{h}.$ Now applying L'Hopital 's Rule and using the continuity of $f'$ we get $L=\lim_{h\to 0 }(f'(3+h)+f'(3-h))= 2 f'(3)=2$

$\begin{aligned} L & = \lim_{h \to 0} \frac 1h \int_{3-h}^{3+h} \frac {4x-2}{x^2+1} dx \\ & = \lim_{h \to 0} \frac 1h \int_{\color{#D61F06}3}^{3+h} \frac {4x-2}{x^2+1} dx + \lim_{h \to 0} \frac 1h \int^{\color{#D61F06}3}_{3-h} \frac {4x-2}{x^2+1} dx \\ & = \frac d{dx} \int_{3}^{3+h} \frac {4x-2}{x^2+1} dx + \frac d{dx} \int^{3}_{3-h} \frac {4x-2}{x^2+1} dx \\ & = \frac {4x-2}{x^2+1}\bigg|_{x=3} + \frac {4x-2}{x^2+1}\bigg|_{x=3} \\ & = 1 + 1 = \boxed{2} \end{aligned}$

This question can be easily solved by using two important methods namely, use of L'hospital rule for limits and another one being DUIS i.e. Differentiation Under Integral Sign. As we have integration with limits 3+h and 3-h with h tending to 0, so numerical value of integration tends to 0, So this is case of 0/0, and so to evaluate the value of limits we can use L'hospital rule, which then requires DUIS in the numerator(need not to say that denominator becomes 1 after differentiation).