Find $\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$ Find the limit of $$\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$$ without l'Hospital.
I tried expanding $1-\cos(x)$ to get something this kind $$\lim_{x\to 0}\frac{\sin x}{x}$$ but couldn't solve it. The answer is $1/p$.
 A: A little bit of trigonometry does the job:
$$\begin{align}
\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)} &= \lim_{x\to 0}\frac{1+\sqrt 2\sin(x-\pi/4)}{1+\sqrt 2 \sin(px-\pi/4)} \\[1mm]
&= \lim_{x\to 0}\frac{\frac{1}{\sqrt 2} + \sin(x-\pi/4)}{\frac{1}{\sqrt 2} + \sin(px - \pi/4)} \\[1mm]
&= \lim_{x\to 0}\frac{\sin(\pi/4) + \sin(x-\pi/4)}{\sin(\pi/4) + \sin(px - \pi/4)} \\[1mm]
&= \lim_{x\to0} \frac{\sin(x/2)\cos(\pi/2)}{\sin(px/2)\cos(\pi/2)} \\[1mm]
&= \lim_{x\to0} \underbrace{\frac{\sin(x/2)}{\color{red}{x/2}}}_{\to 1}\underbrace{\frac{\color{blue}p\color{red}{x/2}}{\sin(px/2)}}_{\to 1}\frac {1}{\color{blue}p} \\[1mm]
&= \frac 1p\end{align}$$

See MSE discussions about $\lim_{x\to0}\frac{\sin x}{x}=1$

A: Let $f(x) = \sin{x} - \cos{x}$. Then, $$1 = f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{\sin{h} - \cos{h} + 1}{h}$$ Make the substitution $h\mapsto ph$ and this limit becomes $$1 = f'(0) = \lim_{h \to 0} \frac{\sin{ph} - \cos{ph} + 1}{ph}$$ Finally, note that: $$ 1 = \frac{1}{1} =  \frac{f'(0)}{f'(0)} =  \lim_{h \to 0} \left ( \frac{\frac{\sin{h} - \cos{h} + 1}{h}}{\frac{\sin{ph} - \cos{ph} + 1}{ph}} \right )$$ So, $$\lim_{h\to 0} \frac{1+\sin h-\cos h}{1+\sin{ph}-\cos{ph}} = \frac{1}{p}$$
A: $$\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}=\lim_{x\to 0} \frac{\sin x+2\sin^2 x/2}{\sin(px)+2\sin px/2}.$$
Make use of the properties of the limit:
$$ \frac{\lim_{x\to 0}\sin x+\lim_{x\to 0}2\sin^2 x/2}{\lim_{x\to 0}\sin(px)+\lim_{x\to 0}2\sin px/2} = \lim_{x\to 0}\frac{\sin x}{\sin px}.$$
Algebraic manipulation:
$$\lim_{x\to 0}\frac{x\sin x}{x\sin px}=\lim_{x\to 0}\frac{\sin x}{x}\lim_{x\to 0}\frac{x}{\sin px}=1\cdot\lim_{x\to 0/p}\frac{x/p}{\sin x}=\frac{1}{p}.$$
Here is a geometric approach (without l'hopital) that proves $\lim_{x\to 0}\frac{\sin x}{x}=1.$
A: With some basic asymptotic analysis;
First, observe that $\:1-\cos x+\sin x=2\sin^2 \frac x2$.
Now, $\sin^2\frac x2=o(\sin x)$ near $0$, and therefore
$$1-\cos x+\sin x\sim_0\sin x \sim_0x.$$
Similarly, $\;1-\cos px+\sin px\sim_0\sim px\sim_0px$, whence
$$\frac{1-\cos x+\sin x}{1-\cos px+\sin px}\sim_0\frac x{px}=\frac1p.$$
A: Substitute $x=2t$, so
$$
1+\sin x-\cos x=1+2\sin t\cos t-1+2\sin^2t=2\sin t(\sin t+\cos t)
$$
and your limit becomes
$$
\lim_{t\to0}\frac{2\sin t(\sin t+\cos t)}{2\sin(pt)(\sin(pt)+\cos(pt))}=
\lim_{t\to0}\frac{1}{p}\frac{\sin t}{t}\frac{pt}{\sin(pt)}\frac{\sin t+\cos t}{\sin(pt)+\cos(pt)}
$$
