Question

Show why the limit as x approaches 0 (csc(x)-cot(x)) involves an indeterminate form, and then prove that the limit equals 0.

Answer 1

Answer with Step-by-step explanation:

We are given that $\lim_{x\rightarrow 0 }(csc(x)-cot(x))$

We have to prove that why the limit x approaches 0(csc(x)-cot(x)) involves an indeterminate form and prove that the limit equals to 0.

$\lim_{x\rightarrow 0 }(\frac{1}{sinx}-\frac{cosx}{sinx})$

Because $csc(x)=\frac{1}{sinx}$ and$cot(x)=\frac{cosx}{sinx}$

$\lim_{x\rightarrow 0 }(\frac{1-cosx}{sinx})$

$\frac{1-cos0}{sin0}$

We know that cos 0=1 and sin 0=0

Substitute the values then we get

$\frac{1-1}{0}=\frac{0}{0}$

We know that $\frac{0}{0}$ is indeterminate form

Hence, the limit x approaches 0(csc(x)-cot(x)) involves an indeterminate form.

L'hospital rule:Apply this rule and  differentiate numerator and denominator separately when after applying $\lim_{x\rightarrow a }$we get indeterminate form$\frac{0}{0}$

Now,using L' hospital rule

$\lim_{x\rightarrow 0 }\frac{0+sinx}{cosx}$

because $\frac{dsinx}{dx}=cosx,\frac{dcosx}{dx}=-sinx}$

Now, we get

$\lim_{x\rightarrow 0 }\frac{sinx}{cos x}$

$\frac{sin0}{cos0}$

$\frac{0}{1}=0$

Hence,$\lim_{x\rightarrow 0 }(csc(x)-cot(x))=0$