Doubt from limits

solved
limits
trigonometry
find-limit
easy
pp
#1

Question no 59 i didn't understand in last step

#2

i think something missing better use l hospital rule.. @abhijit_2020

#3

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#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}

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#5

Question no 75 i didnt understand how root of cos inverse x became root of pi in third step

#6

Question no 7 and 9

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#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.

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#8

@abhijit_2020

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

#9

@abhijit_2020

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.

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#10

@abhijit_2020

For Q.7

Do let me know if you have any doubts.

#11

Sir the solution of question 7 is not clearly visible

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#12

Thanks for telling @abhijit_2020
I shall repost it, sorry for the inconvenience. :slight_smile:
Do mark the posts as solutions if you find them satisfying, thanks!
@abhijit_2020

#13

@abhijit_2020

For Q.7

Is this fine? Do let me know if you have any doubts.

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#14

Now its fine

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#15

How to do question 6 and 7 by l hospital method

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#16

@abhijit_2020

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.

#17

@abhijit_2020

For Q.6

Do let me know if you have any doubts.

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#18

For Q.7(a)

Do let me know if you have any doubts.

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#19

@abhijit_2020

For Q.7(b)

Do let me know if you have any doubts.

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#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

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