Stuck on a trig limit

**xxlvh** · September 19th 2009, 08:48 PM

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

I was never any good at these. >.<

**Prove It** · September 19th 2009, 08:58 PM

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

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

**xxlvh** · September 19th 2009, 09:00 PM

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.

**Prove It** · September 19th 2009, 09:10 PM

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.

**courteous** · September 19th 2009, 09:10 PM

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

**xxlvh** · September 19th 2009, 09:14 PM

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?

**Prove It** · September 19th 2009, 09:20 PM

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 $\frac{2}{0}$ .

**mr fantastic** · September 20th 2009, 01:46 AM

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.

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