Question no 59 i didn't understand in last step

# Doubt from limits

**abhijit_2020**#1

**Hari_Shankar**#4

Here is an interesting approach once you know a limit exists:

Let \displaystyle L= \lim_{x \rightarrow 0} \left( \dfrac{1}{x^2} -\dfrac{1}{\sin^2 x} \right)

Replacing x with 2x

\displaystyle L= \lim_{x \rightarrow 0} \left( \dfrac{1}{4x^2} -\dfrac{1}{\sin^2 2x} \right)

\displaystyle = \lim_{x \rightarrow 0} \left( \dfrac{1}{4x^2} -\dfrac{1}{4\sin^2 x \cos^2 x} \right)

\displaystyle =\dfrac{1}{4} \lim_{x \rightarrow 0} \dfrac{\sin^2 x \cos^2x - x^2}{x^2 \sin^2 x \cos^2 x}= \dfrac{1}{4} \lim_{x \rightarrow 0} \dfrac{\sin^2 x - x^2-\sin^4 x}{x^2 \sin^2 x \cos^2 x}

Now \ \displaystyle \lim_{x \rightarrow 0} \dfrac{\sin^2 x - x^2}{x^2 \sin^2 x \cos^2 x} - \dfrac{\sin^4 x}{ x^2 \sin^2 x \cos^2 x} = L-1

So we get L= \dfrac{L-1}{4} so that L = -\dfrac{1}{3}

**abhijit_2020**#5

**Dhathri_1**#7

It is shown that X tends to -1...which implies cos inverse x tends to π. That is why it is replaced with π.

What are the answers to Q.7 and Q.9? Please do share the answers.

Okay got it.

For Q.9

This kinda limit is called as *Path Limits* and are evaluated just as we evaluate path integrals, etc.

Do let me know if you have any doubts.

**Churchil_2018**#12

Thanks for telling @abhijit_2020

I shall repost it, sorry for the inconvenience.

Do mark the posts as solutions if you find them satisfying, thanks!

@abhijit_2020

**Churchil_2018**#13

**Churchil_2018**#16

Before applying the L Hospital rule, you should be know where is it applicable. It is applicable only for the forms \frac{0}{0} or \frac{\infty}{\infty}. And hence, first you need to transform the problem into one of that kind then only you can apply the rule. I shall try out your problems and post the solutions.

**Hari_Shankar**#20

You can also use the result you got in 7(a)

\displaystyle \ln x^{x^x} = x \ln x^x \rightarrow 0 as x \rightarrow 0, \ \because x^x \rightarrow 1 as x \rightarrow 0