Limit with sequence of functions and integrals

Question

I need some help with a question. I have to calculate

$$\lim_{x \to 3}\frac{x^2}{x - 3}\int_3^x \frac{\sin t}{t}dt.$$

If I'm not wrong, we can write

$$\sin(x) = \sum_{n = 0}^{\infty}\frac{x^{2n+1}}{(2n+1)!} \Longrightarrow \frac{\sin(x)}{x}=\sum_{n = 0}^{\infty}\frac{x^{2n}}{(2n+1)!},$$

then

$$\int_3^x \frac{\sin (t)}{t}dt = \sum_{n = 0}^{\infty}\int_3^x \frac{t^{2n}}{(2n+1)!}dt = \sum_{n = 0}^{\infty}\frac{t^{2n+1}}{(2n+1)(2n+1)!}\Bigg|_3^x.$$

But I don't know how exactly to use it. I appreciate any help!

Answer 1

By L'Hospital's rule and the fundamental theorem of calculus:

$$\lim_{x\to 3} \frac{x^2}{x-3} \int_3^x \frac{\sin t}{t} dt = 9\lim_{x\to 3}\frac{\int_3^x \frac{\sin t}{t} dt}{x-3} = 9\lim_{x\to 3} \frac{\sin x}{x} = 3\sin 3$$

Answer 2

You can use L'Hospital's rule and the fundamental theorem: $$ \lim\limits_{x\to 3}(...)=\lim\limits_{x\to 3}\frac{2x\int_3^x\frac{\sin t}{t}\,dt+x^2\frac{\sin x}{x}}{1}=3\sin 3. $$

Answer 3

Using L’Hospital Rule and the Fundamental Theorem of Calculus, we have

$$ \begin{aligned} & \lim _{x \rightarrow 3} \frac{x^{2}}{x-3} \int_{3}^{x} \frac{\sin t}{t} d t \quad\left(\frac{0}{0}\right) \\ =& \lim _{x \rightarrow 3} \frac{x^{2} \frac{\sin x}{x}+2 x \int_{3}^{x} \frac{\sin t}{t} d t}{1} \\ =& \frac{3^{2} \sin 3}{3}+0 \\ =& \frac{\sin 3}{3} \end{aligned} $$