# Use algebraic methods to evalute each limit (if it exists).` lim_(x-gt3) (x-3)/(sqrt(x+6) -3)` and `lim_(x-gt-3) ((1/x) + (1/3))/(x+3)`

beckden | High School Teacher | (Level 1) Educator

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`lim_(x->3) (x-3)/(sqrt(x+6)-3)` we are going to multiply top and bottom by (sqrt(x+6)+3)

We do this because `(x-a)(x+a) = x^2 - b^2`

we get

`=lim_(x->3) (x-3)/(sqrt(x+6)-3) * (sqrt(x+6)+3)/(sqrt(x+6)+3)` we get

`=lim_(x->3) ((x-3)(sqrt(x+6)+3))/((sqrt(x+6)-3)(sqrt(x+6)+3))`

`=lim_(x->3) ((x-3)(sqrt(x+6)+3))/(x+6-3)`

This equation has no problem evaluating at x=3 so we get

`lim_(x-gt3)(x-3)/(sqrt(x+6)-3) = ((3-3)(sqrt(3+6)+3))/(3+6-3) = ((0)(6))/(6) = 0`

For the second limit

lim_(x->-3)(1/x + 1/3)/(x+3)

We can simplify `1/x+1/3 = 3/(3x) + x/(3x) = (x+3)/(3x)`

So

`lim_(x->-3) (1/x+1/3)/(x+3) = lim_(x->-3) ((x+3)/(3x))/(x+3) = lim_(x->-3) (x+3)/(3x) * 1/(x+3) = lim_(x->-3) 1/(3x)`

Now we have no problem evaluating this expression

`lim_(x->-3) (1/x + 1/3)/(x+3) = 1/(3(-3)) = -1/9`