Thread: Stuck on a trig limit

1. Stuck on a trig limit

$\lim_{x\to\\1^+}\frac{x+1}{x\sin{{\pi}x}}$

I was never any good at these. >.<

2. Originally Posted by xxlvh
$\lim_{x\to\\1^+}\frac{x+1}{x\sin{{\pi}x}}$

I was never any good at these. >.<
Surely since $\sin{\pi x} \to 0$ as $x \to 1$, and $x + 1 \to 2$, then

$\frac{x + 1}{x\sin{\pi x}} \to \frac{2}{0} \to \infty$...

3. Is there no possible way to rearrange it? I was told to always try that when a denominator went to 0, which of course I can't seem to do on this question..
Also the answer is negative infinity, not positive.

4. Originally Posted by xxlvh
Is there no possible way to rearrange it? I was told to always try that when a denominator went to 0, which of course I can't seem to do on this question..
Also the answer is negative infinity, not positive.
I didn't notice the "right hand limit" part.

There are several ways to rearrange the expression, including

$\frac{x + 1}{x \sin{\pi x}} = \frac{x}{x\sin{\pi x}} + \frac{1}{x\sin{\pi x}}$

$= \frac{1}{\sin{\pi x}} + \frac{1}{x \sin{\pi x}}$

$= \frac{1}{\sin{\pi x}}\left(1 + \frac{1}{x}\right)$.

All will lead to the same conclusion though.

5. You could do l'Hospital's rule and get: $\frac{1}{x\pi\cos(\pi x)+\sin(\pi x)} \rightarrow -\frac{1}{\pi}$

6. I don't believe L'Hopitals is applicable in this case, the numberator does not go to 0.

Hmm, is there any way of obtaining the answer of negative infinity algebraically then?

7. Originally Posted by courteous
You could do l'Hospital's rule and get: $\frac{1}{x\pi\cos(\pi x)+\sin(\pi x)} \rightarrow -\frac{1}{\pi}$
You most definitely can NOT use L'Hospital's Rule in this case.

To use L'Hospital, the expression must tend to an indefinite form, i.e. $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

But like I showed earlier, the expression tends to $\frac{2}{0}$.

8. Originally Posted by xxlvh
[snip]Also the answer is negative infinity, not positive.
Note that $\lim_{x \rightarrow 1^+} \sin (\pi x) = 0^-$ since the approach is from the third quadrant.