Please my doubt . Also pls suggest any graphical way to solve this out . It's urgent . Thankyou

The Graph  of this function can be plotted using online tool 

Search desmos in google

Unfortunately this website does not allow attachments for answers hence cannot post how graph looks like.

As you can see arcsin is only defined in -1 to 1.

To find the limit of this function it is sufficient to find LHL whhich from the graph comes out to be 1

If you wnat to learn more about plotting such graphs yourself refer to differential calculus book by Amit Agarwal from Arhant.

This book does a good job of explaining  many problems using graphs.

sir but why are we considering LHL from the graph? why not RHL?  also thanks i found graph for this function. and sir is cengage series maths alone sufficient for jee?

 since arcsin(x) is defined only on [-1,1] you can find only LHL not RHL as you cant approach the function with x>1 (hope this clears your doubt)

 

Yes cengage  learning book is very good and sufficient for jee

But still you should try to practise vast variety of questions which surely cannot be given in a single book

 

You re welcome neha. Regarding books, Cengage has good reviews online, but I have not personally refered to it so cant comment on sufficiency of that book.

I hope cengage book  should have explained the kind of problems you have asked 

I refer to arihant books which explains many calculus problems graphically and also have a chapter where it explains how to draw graphs of such functions.

Another thing is that its ok to refer to single book provided it  covers all kind of questions asked in JEE and you are able to complete all questions.  But there is no harm in trying other books if you are finding one particular book insufficient . 

 

 

Thanks sir for ur suggestion . Also I have one doubt regarding that question. Why can't we have x>1 when we have X

I hope you want to ask why domain of arcsin cannot be greater than 1. 

 by definition of inverse function the  the domain of arcsin = range of sin function.

Now range of sin function is [-1, 1 ]

 

Another way to look at this as follows.

Assume arcsin exists for x >1 or y= arcsin(x) 

it means there exist an angle y for which

               sin y =  x  where x is greater than 1

but max value of siny is 1 hence our assumption is wrong.