1. ## Limit Problem

It looks like it should be simple, but I'm having some trouble wrapping my head around this one. Any help would be greatly appreciated.

$\displaystyle \lim_{x\rightarrow3}\bigg(\frac{x}{x-3}\int_{3}^{x}\frac{\sin{t}}{t} dt\bigg)$

Thanks.

2. Originally Posted by McScruffy
It looks like it should be simple, but I'm having some trouble wrapping my head around this one. Any help would be greatly appreciated.

$\displaystyle \lim_{x\rightarrow3}\bigg(\frac{x}{x-3}\int_{3}^{x}\frac{\sin{t}}{t} dt\bigg)$

Thanks.
Using the usual algebra of limits you have $\displaystyle \left( \lim_{x\rightarrow3} x\right) \cdot \lim_{x \rightarrow 3} \frac{\int_{3}^{x}\frac{\sin{t}}{t} dt}{x - 3}$.

The second limit has the indeterminant form 0/0 so apply l'Hopitals to it (use the Fundamental Theorem of Calculus to differentiate the numerator).

3. Originally Posted by mr fantastic
Using the usual algebra of limits you have $\displaystyle \left( \lim_{x\rightarrow3} x\right) \cdot \lim_{x \rightarrow 3} \frac{\int_{3}^{x}\frac{\sin{t}}{t} dt}{x - 3}$.

The second limit has the indeterminant form 0/0 so apply l'Hopitals to it (use the Fundamental Theorem of Calculus to differentiate the numerator).
Got it, thanks.

4. Originally Posted by McScruffy
It looks like it should be simple, but I'm having some trouble wrapping my head around this one. Any help would be greatly appreciated.

$\displaystyle \lim_{x\rightarrow3}\bigg(\frac{x}{x-3}\int_{3}^{x}\frac{\sin{t}}{t} dt\bigg)$

Thanks.
Perhaps you could write $\displaystyle \int_{3}^{x}\frac{\sin{t}}{t} dt$ as a function of x.

That is $\displaystyle F(x)=\int_{3}^{x}\frac{\sin{t}}{t}dt$

Not really sure how to proceed though...

5. Let F(t) be an antiderivative of $\displaystyle \frac{\sin t}{t}$

then $\displaystyle \lim_{x \to 3} \frac{x}{x-3} \int^{x}_{3} \frac{\sin t}{t} \ dt = \lim_{x \to 3} \frac{x}{x-3} \big(F(x)-F(3)\big)$

$\displaystyle = \lim_{x \to 3} x \cdot \lim_{x \to 3} \frac{F(x)-F(3)}{x-3}$

$\displaystyle = 3F'(3) = 3 \ \frac{\sin 3}{3} = \sin 3$

6. Originally Posted by Random Variable
Let F(t) be an antiderivative of $\displaystyle \frac{\sin t}{t}$

then $\displaystyle \lim_{x \to 3} \frac{x}{x-3} \int^{x}_{3} \frac{\sin t}{t} \ dt = \lim_{x \to 3} \frac{x}{x-3} \big(F(x)-F(3)\big)$

$\displaystyle = \lim_{x \to 3} x \cdot \lim_{x \to 3} \frac{F(x)-F(3)}{x-3}$

$\displaystyle = 3F'(3) = 3 \ \frac{\sin 3}{3} = \sin 3$
That was really nice, man.