Thread: Could someone check my answers for limits please?

Could someone check my calculus answers, 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!

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:

. . . . .

"Cancelling" gives us:

. . . . .

Evaluating at x = 3 results in:

. . . . .

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:

. . . . .

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

. . . . .

. . . . .

. . . . .

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