# Calculating indeterminate form limits involving $\cos(x)$ and $\sin(x)$, using only algebraic manipulation

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.

• The key is making use of the result $\lim_{x\to 0}\frac{\sin x}{x}=1.$ – Yuta May 1 '19 at 3:52
• @Yuta I thought so, but what shall I do if there is $cos(x)$? – AaronTBM May 1 '19 at 3:58
• 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). – Blue May 1 '19 at 4:00
• Don’t forget your cofunction identities. $cos(x)=sin(x+\frac{\pi}{2})$. – H Huang May 1 '19 at 4:01
• 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. – 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.