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I came across this limit:

$$\lim\limits_{x\to1}{\dfrac{1-x}{1-\sin\left(\dfrac{\pi \cdot x}{2}\right)}}$$

Now when I evaluated it myself I simply got $\infty$ applying L'Hospital's rule in this way:

$$\lim\limits_{x\to1}{\dfrac{2}{\pi \cdot \cos\left(\dfrac{\pi \cdot x}{2}\right)}} = \frac{2}{0} = \infty$$

When I went online to check if I got the right solution, the evaluator said that the limit diverges, and that when we approach it from $1^+$ we get $-\infty$, and when we approach it from $1^-$ we get $\infty$. There is even a graph under the evaluation, really showing that the result is correct. My question is, could anyone please shed some light on this to me please?

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  • $\begingroup$ You cannot distribute limits over a quotient when the bottom is zero. $\endgroup$ – Seth Jun 1 '14 at 21:56
  • $\begingroup$ You have to consider when cos is positive and when it is negative. $\endgroup$ – Seth Jun 1 '14 at 21:56
  • $\begingroup$ As you approach from the left, cos is positive so the limit is infinitiy, while from the right it is negative, so the limit is negative infinity. $\endgroup$ – Seth Jun 1 '14 at 21:58
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From the left side, you see that the limit goes to $+\infty$: $$ \begin{array}{c|lcr} x & f(x) \\ \hline 0 & 0.6366 \\ 0.5 & 0.9003 \\ 0.8 & 2.0601 \\ 0.9 & 4.0696\\ 0.95 & 8.114034\\ 0.99 & 40.53014 \\ \end{array} $$

However, from the right side, it approaches $-\infty$: $$ \begin{array}{c|lcr} x & f(x) \\ \hline 2 & -0.6366 \\ 1.5 & -0.9003 \\ 1.2 & -2.0601 \\ 1.1 & -4.0696\\ 1.05 & -8.114034\\ 1.01 & -40.53014 \\ \end{array} $$

Because $\lim_{x\to 1^+} \ne \lim_{x\to1^-}$, $\lim_{x\to 1}$ doesn't exist.

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$\cos{x}< 0$ for $x \in \left(\dfrac{\pi}{2},\pi\right)$, and $\cos{x}> 0$ for $x \in \left(0,\dfrac{\pi}{2}\right)$

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The denominator is strictly positive. The numerator changes signs depending on which direction you approach from.

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