Question

If \[\LARGE{\lim_{x \rightarrow 1}\frac{asin(x-1)+bcos(x-1)+4}{x^2-1}=-2}\], then a-b=??

Answer 1

This is a zero-by-zero limit, so you can apply L'Hopital's Rule if you want to.

Answer 2

\[\lim_{x \rightarrow 1}\frac{acos(x-1)-bsin(x-1)+4}{2x}\]am i correct??

Answer 3

Wait, why did I call it a zero-by-zero limit -_-

Answer 4

since the indeterminate form 0/0 is coming on putting x=1

Answer 5

But for the limit to exist, yeah, the numerator must be zero. So \(b = -4\).

Answer 6

And then you apply L'Hopital's rule.

Answer 7

sorry but i don't gt why the numerator must be zero

Answer 8

In order for a limit to exist there shud be a zero in numerator otherwise the limit wouldnot be defined since there is a 0 in denominator

Answer 9

yes since RHS is a finite quantity, at x->1 numerator must be 0
so
0+b+4=0
b=-4

and then try putting x=1+p and simplify

Answer 10

^^^

To sum up, we need two conditions:

1. The limit should be 0/0-type.
2. The limit should be -2 using l'Hopital's rule.

Answer 11

a will also come out to be -4 so ur ans wud be 0

Answer 12

thx :)