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 )((1)/(sinx)-(cosx)/(sinx))`

Because
`csc(x)=(1)/(sinx)` and
`cot(x)=(cosx)/(sinx)`

`\lim_(x\rightarrow 0 )((1-cosx)/(sinx))`

`(1-cos0)/(sin0)`

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

Substitute the values then we get

`(1-1)/(0)=(0)/(0)`

We know that
`(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
`(0)/(0)`

Now,using L' hospital rule

`\lim_(x\rightarrow 0 )(0+sinx)/(cosx)`

because
`(dsinx)/(dx)=cosx,(dcosx)/(dx)=-sinx}`

Now, we get

`\lim_(x\rightarrow 0 )(sinx)/(cos x)`

`(sin0)/(cos0)`

`(0)/(1)=0`

Hence,
`\lim_(x\rightarrow 0 )(csc(x)-cot(x))=0`