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Find the limit $$ \lim_{x\to 3}\left(\frac{x}{x-3}\int_{3}^{x}\frac{\sin t}{t} dt\right) $$ without using L'Hopital's rule.

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By the Mean Value Theorem, $$\int_3^x\frac{\sin t}{t}=(x-3) \frac{\sin c_x}{c_x}$$ for some $c_x$ between $3$ and $x$. So our product is equal to $$x\cdot\frac{\sin c_x}{c_x}.$$ As $x\to 3$, $\sin{c_x}\to\sin 3$ and $c_x\to 3$, so our limit is $\sin 3$.

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Limit definition of the derivative:

$$f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$$

Using this for your situation we set:

$$f(x) = x\int_3^x \frac{\sin(t)}{t} dt$$

$$a = 3$$

and arrive at the limit you want to evaluate since $f(3) = 0$ due to the integral. So we want to find the derivative of the function and evaluate it at $x = 3$:

$$f'(x) = \frac{d}{dx} \Big[ x\int_3^x \frac{\sin(t)}{t} dt \Big]$$

$$= \int_3^x \frac{\sin(t)}{t} dt + x\Big[ \frac{d}{dx}\int_3^x \frac{\sin(t)}{t} dt \Big]$$

$$= \int_3^x \frac{\sin(t)}{t} dt + \sin(x)$$

Now evaluation gives:

$$f'(3) = \int_3^3 \frac{\sin(t)}{t} dt + \sin(3) = \sin(3)$$

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$$L=\lim_{x\to 3}(\frac{x}{x-3}\int_{3}^{x}\frac{\sin t}{t})$$ $$L=\lim_{x\to 3}(1+\frac{3}{x-3})\int_{3}^{x}\frac{\sin t}{t}$$ $$L=\lim_{x\to 3}\int_{3}^{x}\frac{\sin t}{t}+\lim_{x\to 3}\frac{3}{x-3}\int_{3}^{x}\frac{\sin t}{t}$$

$$L=0+3\frac{d}{dx} \left[ \int_{3}^{x}\frac{\sin t}{t} \right]_{x=3} $$

by the fundamental theorem

$$ L= 3\left[ \frac{\sin 3}{3} \right] = \sin 3 $$

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