# Limit $\lim_{x \to 0} \left(\cot x-\frac{1}{\sin x}\right)$

Im having trouble calculating this limit:

$$\lim_{x \to 0} \left(\cot x-\frac{1}{\sin x}\right)$$

I've tried factoring out $$\frac{1}{\sin x}$$, $$\cos x$$, $$\cot x$$ and it doesn't lead me anywhere. I also tried looking at alternative form $$\frac{\cos x-1}{\sin x}$$, with no luck. I can't use l'Hospitals rule. Can anyone help? I think this is super easy but somehow I'm stuck.

$$\frac{(\cos x-1)(\cos x +1)}{\sin x (\cos x+1)}= \frac{\cos^2-1}{\sin x(\cos x +1)}=$$

$$\frac{-\sin^2 x}{\sin x(\cos x+1)}=\frac{-\sin x}{\cos x +1}$$.

And now?

• nice way without using standard limits, thank you – Comfy Cat Apr 4 '20 at 16:52
• Comfy.A pleasure.:) – Peter Szilas Apr 4 '20 at 17:04

Using standard limits

$$\cot x - \frac 1{\sin x} = x\frac{\cos x - 1}{x^2}\frac{x}{\sin x}\stackrel{x\to 0}{\longrightarrow}0\cdot \frac{-1}{2}\cdot 1=0$$

• I forgot about these, thanks – Comfy Cat Apr 4 '20 at 16:52

$$\cot x-\frac{1}{\sin x}=\frac{\cos x}{\sin x}-\frac{1}{\sin x}=\frac{\cos x-1}{\sin x}$$

The Taylor series expansion for $$\sin(x)$$ and $$\cos(x)$$ at $$x=0$$ is

$$\sin(x)=x-\frac{x^3}{3!}+\mathcal{O}(x^5)$$ $$\cos(x)=1-\frac{x^2}{2!}+\mathcal{O}(x^4)$$ therefore $$\cos(x)-1=-\frac{x^2}{2!}+\mathcal{O}(x^4)$$ hence $$\frac{\cos x-1}{\sin x}=\frac{-\frac{x^2}{2!}+\mathcal{O}(x^4)}{x-\frac{x^3}{3!}+\mathcal{O}(x^5)}=\frac{-\frac{x}{2!}+\mathcal{O}(x^3)}{1-\frac{x^2}{3!}+\mathcal{O}(x^4)}$$

then $$\frac{\cos x-1}{\sin x} \to \frac{0}{1}=0$$ as $$x\to 0$$.

$$\lim_{x \to 0} \left(\cot x-\frac{1}{\sin x}\right)=\lim_{x\to 0}\frac{\cos x-1}{\sin x}=\lim_{x\to 0}\frac{-2\sin^2\left(\frac{x}{2}\right)}{2\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}=0.$$