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I'm trying to solve this limit:

$$\lim_{x \to 1}{\frac{1-x^2}{\sin (\pi x)}}$$

The answer ought to be $\frac{2}{\pi}$, but I end up with $0$:

$\lim\limits_{x \to 1}{\frac{1-x^2}{\sin (\pi x)}} = $ $\lim\limits_{y \to 0}{\frac{1-(y+1)^2}{\sin (\pi (y+1))}} = $ $\lim\limits_{y \to 0}{\frac{\pi(y+1)}{\sin (\pi (y+1))} \frac{1-(y+1)^2}{\pi(y+1)}} = $ $\lim\limits_{y \to 0}{\frac{1-(y+1)^2}{\pi(y+1)}} = 0$

Where and why is my solution incorrect?

Note: I'm aware of this post, however I believe mine is different because I'm asking where and why my solution went wrong, not why my answer was wrong.

  • 1
    The step where you have $\lim_{y \rightarrow 0} \frac{\pi (y+1)}{\sin(\pi(y+1)}$ and say that limit is equal to $1,$ how did you get that ? – 2017-02-15
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    I don't see how you can drop the π(y+1) / sin(π(y+1)) in the final step. sin(pi) goes to zero which you may not divide by, so you need to use L'Hôpital's method from the get-go. – 2017-02-15
  • 2
    First write $\sin(\pi(y+1))=-\sin\pi y$ then add in numerator and denominator . – 2017-02-15
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    @MyGlasses -- Will try. Thanks! – 2017-02-15
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    @MyGlasses That is slick! I will have to remember that little nugget. – 2017-02-15

2 Answers 2

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Your third equality attempts to make use of the rule $\lim\limits_{x\to0}\frac{x}{\sin x} = 1$, but note that yours has $y\to 0$ yet the argument is not $y$, it is $\pi(y+1)$, which does not go to zero. That's where your work goes wrong.

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    Oh! I see. I'll see how I can fix it, and come back later if more help is needed. Thanks in the meantime! – 2017-02-15
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    @SirJony You are welcome. Your idea is very clever, even though it didn't work. MyGlasses seems to have a good workaround idea in the comments above, though if it were me, I would just use L'Hospital's rule, if that is allowed. – 2017-02-15
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$$\lim _{ y\to 0 }{ \frac { \pi (y+1) }{ \sin (\pi (y+1)) } } =\frac { \pi }{ 0 } \neq 1\\ $$

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    How come? I thought $\lim\limits_{x\to 0}{\frac{\sin x}{x}} = 1$ (which is the identity I used). – 2017-02-15
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    @SirJony see my answer. Your identity is correct, but the *argument* of the sine function needs to go to zero, which yours does not. – 2017-02-15