Answer with Step-by-step explanation:
We are given that [tex]\lim_{x\rightarrow 0 }(csc(x)-cot(x))[/tex]
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.
[tex]\lim_{x\rightarrow 0 }(\frac{1}{sinx}-\frac{cosx}{sinx})[/tex]
Because [tex]csc(x)=\frac{1}{sinx}[/tex] and[tex]cot(x)=\frac{cosx}{sinx}[/tex]
[tex]\lim_{x\rightarrow 0 }(\frac{1-cosx}{sinx})[/tex]
[tex]\frac{1-cos0}{sin0}[/tex]
We know that cos 0=1 and sin 0=0
Substitute the values then we get
[tex]\frac{1-1}{0}=\frac{0}{0}[/tex]
We know that [tex]\frac{0}{0}[/tex] 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 [tex]\lim_{x\rightarrow a }[/tex]we get indeterminate form[tex]\frac{0}{0}[/tex]
Now,using L' hospital rule
[tex]\lim_{x\rightarrow 0 }\frac{0+sinx}{cosx}[/tex]
because [tex]\frac{dsinx}{dx}=cosx,\frac{dcosx}{dx}=-sinx}[/tex]
Now, we get
[tex]\lim_{x\rightarrow 0 }\frac{sinx}{cos x}[/tex]
[tex]\frac{sin0}{cos0}[/tex]
[tex]\frac{0}{1}=0[/tex]
Hence,[tex]\lim_{x\rightarrow 0 }(csc(x)-cot(x))=0[/tex]
