What is `lim_(x->3) sqrt(x^2 - 9)/(x-3)`

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The value of `lim_(x->3) (x^2 - 9)/(x-3)` has to be determined.

`lim_(x->3) (x^2 - 9)/(x-3)`

= `lim_(x->3) ((x - 3)(x + 3))/(x-3)`

= `lim_(x->3) (x + 3)`

= 6

**The required limit `lim_(x->3) (x^2 - 9)/(x-3) = 6` **

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The limit `lim_(x->3) sqrt(x^2 - 9)/(x - 3)` can be determined in the following way.

`lim_(x->3) sqrt(x^2 - 9)/(x - 3)`

= `lim_(x->3) sqrt((x - 3)(x+3))/(x - 3)`

= `lim_(x->3) sqrt(x+3)/sqrt(x - 3)`

If x = 3, the result is `sqrt 6/0 = oo` .

**The limit **`lim_(x->3) sqrt(x^2 - 9)/(x - 3) = oo`

I'm asking for Square root of x^2-9..

The limit `lim_(x->3) sqrt(x^2 - 9)/(x - 3)` has to be determined.

If we substitute x = 3, the numerator is `sqrt(3^2 - 9)` = 0 and the denominator is 3 - 3 = 0. This gives the indeterminate form `0/0` .

We can use l'Hopital's rule and substitute the numerator and denominator with their derivatives.

This gives:

`lim_(x->3) ((1/2)*2*x*1/(sqrt(x^2 - 9)))/1`

Substituting x = 3 gives:

`((1/2)*2*3*1/(sqrt(3^2 - 9)))/1`

= `3/sqrt(0)`

This is equal to infinity.

The limit `lim_(x->3) sqrt(x^2 - 9)/(x - 3) = oo`

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