Question

solution of this one

Answer 1

1.)

In given limit
x tends to 2-

So x is nearer to 2 from left side...

So x - 2 will always be negative...
Thus modulus will be open as negative (x -2)

(x-2) will get cancel in numerator and denominator.

So answer is -1.

2.)

$= \frac{2x + 3}{ \sqrt{ {x}^{2} - 1 } }$

$= \frac{2x + 3}{ x\sqrt{ 1 - \frac{1}{ {x}^{2} } } }$

$= \frac{2 + (\frac{3}{x} )}{ \sqrt{ 1 - \frac{1}{ {x}^{2} } } }$

Putting x tends to infinity we get,

$= \frac{2 + (0)}{ \sqrt{ 1 - 0 } }$

$= 2$