Question

question
evaluate the limit:
lim_{h
ightarrow0}\frac{5csc(x + h)-5csc(x)}{h}

Explanation:

Step1: Rewrite csc in terms of sin

Recall that $\csc(x)=\frac{1}{\sin(x)}$. So the limit becomes $\lim_{h
ightarrow0}\frac{5\frac{1}{\sin(x + h)}-5\frac{1}{\sin(x)}}{h}=\lim_{h
ightarrow0}\frac{5(\sin(x)-\sin(x + h))}{h\sin(x)\sin(x + h)}$.

Step2: Use the trigonometric identity $\sin(A)-\sin(B)=2\cos(\frac{A + B}{2})\sin(\frac{A - B}{2})$

Here $A=x$ and $B=x + h$, so $\sin(x)-\sin(x + h)=2\cos(x+\frac{h}{2})\sin(-\frac{h}{2})=- 2\cos(x+\frac{h}{2})\sin(\frac{h}{2})$. The limit is now $\lim_{h
ightarrow0}\frac{-10\cos(x+\frac{h}{2})\sin(\frac{h}{2})}{h\sin(x)\sin(x + h)}$.

Step3: Use the limit property $\lim_{u

ightarrow0}\frac{\sin(u)}{u}=1$
Let $u = \frac{h}{2}$, then $h = 2u$ and as $h
ightarrow0$, $u
ightarrow0$. The limit becomes $\lim_{h
ightarrow0}\frac{-10\cos(x+\frac{h}{2})\sin(\frac{h}{2})}{2\cdot\frac{h}{2}\sin(x)\sin(x + h)}$. We know that $\lim_{h
ightarrow0}\frac{\sin(\frac{h}{2})}{\frac{h}{2}} = 1$. So the limit is $\frac{- 5\cos(x)}{\sin^{2}(x)}=-5\csc(x)\cot(x)$.

Answer:

$-5\csc(x)\cot(x)$