I was doing some calculus homework and I came across with some problems. I have to find the following limits

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

2) $\displaystyle\lim_{x\to 0} \frac{x\cdot\sin(x)}{1-\cos(x)}$

3) $\displaystyle\lim_{x\to \infty} x\cdot\sin\left(\frac{\pi}{x}\right)$

4) $\displaystyle\lim_{x\to \frac{\pi}{2}} \frac{\cos(x)}{x-\frac{\pi}{2}}$

The thing is that I don't know how to solve them because all the things that I tried led me to an indetermination. I don't have to use derivatives or anything similar, just algebra "tricks". My intention isn´t having my homework done by somebody else, but I can´t come up with any idea.

  • 3
    $\begingroup$ The key is making use of the result $\lim_{x\to 0}\frac{\sin x}{x}=1.$ $\endgroup$ – Yuta May 1 '19 at 3:52
  • $\begingroup$ @Yuta I thought so, but what shall I do if there is $cos(x)$? $\endgroup$ – AaronTBM May 1 '19 at 3:58
  • 2
    $\begingroup$ Two relations between sine and cosine are helpful: $\sin^2x +\cos^2x=1$, and $\cos(x) = \sin(\frac{\pi}{2}-x)$. The former helps with (1) and (2); the latter helps with (4). $\endgroup$ – Blue May 1 '19 at 4:00
  • $\begingroup$ Don’t forget your cofunction identities. $cos(x)=sin(x+\frac{\pi}{2})$. $\endgroup$ – H Huang May 1 '19 at 4:01
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    $\begingroup$ Notice the limit given. For $x\to\frac{\pi}{2}$, the substitution $y=\frac{\pi}{2}-x$ is constructive. For $x\to\infty$, the substitution $y=\frac{1}{x}$ should be considered. $\endgroup$ – Yuta May 1 '19 at 4:02


For (1) and (4), let $u=x-\pi/2$.

For (3), let $u=\pi/x$.

Use the identities $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$ and $\cos\theta=1-2\sin^2(\theta/2)$ if necessary.

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