# Math Help - Could someone check my answers for limits please?

1. Find lim x -> 3 (9-x^2)/(x^2-4x+3) I got -3/2

2. Find lim x -> -1 (sqrt(5+x)-2)/1+x, I got 1/4 (Only the 5+x is under the square root, just to clarify.)

If these are wrong, could someone please tell me the correct answer and how to get it so I know how?

Thanks so much!

2. Originally Posted by anita123
1. Find lim x -> 3 (9-x^2)/(x^2-4x+3) I got -3/2
The rational function factors as:

. . . . . $\frac{(3\, -\, x)(3\, +\, x)}{(x\, -\, 3)(x\, +\, 1)}\, =\, \frac{-1(x\, -\, 3)(x\, +\, 3)}{(x\, -\, 3)(x\, +\, 1)}$

"Cancelling" gives us:

. . . . . $-\frac{x\, +\, 3}{x\, +\, 1}$

Evaluating at x = 3 results in:

. . . . . $-\frac{3\, +\, 3}{3\, +\, 1}\, =\, -\frac{6}{4}\, =\, -\frac{3}{2}$

So this looks good!

Originally Posted by anita123
2. Find lim x -> -1 (sqrt(5+x)-2)/1+x, I got 1/4 (Only the 5+x is under the square root, just to clarify.)
I will guess that all of "1 + x" is in the denominator, so the expression is as follows:

. . . . . $\frac{\sqrt{5\, +\, x}\, -\, 2}{1\, +\, x}$

When dealing with problematic expressions containing radicals, it can often be useful to multiply by the conjugate, so let's try that here:

. . . . . $\left(\frac{\sqrt{5\, +\, x}\, -\, 2}{1\, +\, x}\right)\left(\frac{\sqrt{5\, +\, x}\, +\, 2}{\sqrt{5\, +\, x}\, +\, 2}\right)$

. . . . . $\frac{(5\, +\, x)\, -\, 4}{(1\, +\, x)\left(\sqrt{5\, +\, x}\, +\, 2\right)}$

. . . . . $\frac{1\, +\, x}{(1\, +\, x)\left(\sqrt{5\, +\, x}\, +\, 2\right)}$

"Cancelling" the common factor and then evaluating at x = -1 leads to the result you got. Excellent!