limit f(x) = limit (sqrt(x-3) /(x^2-9) x--> 3

By substitution: limit f(x) = 0/0

Then let us factore the denominator:

limit f(x) = limit (sqrt(x-3) / (x+3)(x-3)

= limit (x-3)^1/2 / (x+3)(x-3)

= limit 1/(x+3)(x-3)^1/2

When x --> 3:

==> limit f(x) = 1/6*0 = 1/0 = +infinity;

Then the limit = +infinity

To evaluate lt (sqrtx-3)/(x^2-9) as x--> 3.

When x= 3, Numerator = sqrt3 - 3. and denominator = 3^2-9 = 0.

So the function at x= 3 does not exist.

Left limit = lt h-->0 of (sqrt(3-h)-3)/ ((9-h) - 9) = -ve/-h as = positive ifinite.

Right limit = Lt h--> 0 {sqrt(3+h) - 3/ (9+h-9) } = -ve/ h = -negative onfinite.

So the Left limit is not equal to right limit . And the function does not exist at x =3. Hence limit as x--> 3 does not exist.

We have to calculate lim [(sqrt x-3)/(x^2-9)].

We'll follow the steps:

-we'll write the difference of squares as:

x^2 - 9 = (x-3)(x+3),

Also, we could consider x - 3 as a difference of squares.

x-3 = (sqrtx-sqrt3)(sqrtx+sqrt3)

We'll evaluate the limit:

lim [(sqrt x-3)/(x^2-9)]=lim [(sqrt x-3)/(sqrtx-sqrt3)(sqrtx+sqrt3)]

We'll substitute x by 3, into the limit:

lim (sqrt3 - 3)/(sqrtx+sqrt3)(sqrt3-sqrt3)

**lim [(sqrt x-3)/(x^2-9)] = (sqrt3 - 3)/0*(sqrtx+sqrt3)**

**lim [(sqrt x-3)/(x^2-9)] = +infinite**