Maths limits The question is: lim of x-->3 for (x/(x^2-9))?

Question

How do you do this question? Please show working out. Thank you.

Answer 1

The expression you wrote has not limit as x goes to 3. Instead, it grows without bound.

What you may have meant to write is the limit of ((x-3)/(x^2-9)).
For that expression, both numerator and denominator go to 0 as x goes to 3. So there is a chance that the expression will have a limit.

It turns out that it does have a limit. In fact, you can find it by first factoring the denominator (x^2 - 9) = (x+3)(x-3), then dividing numerator and denominator by (x-3) (the result is (1/(x+3)), and determining the limit of that expression as x goes to 3 (it equals 1/6).

Note: The first responder (Helmut) gave you the same answer that I did (1/6) by applying l'Hospital's rule. But he did not modify the expression you gave in your question, which I said was required in order to find a limit. But he should not have applied l'Hospital's rule to a a fraction where the denominator goes to 0 at the limit, but the numerator does not. If he had recognized that the numerator has to go to 0, he likely would have done what I did, which is to guess that the numerator should be (x-3), which goes to 0 as x goes to 3. Had he done that, he would have found the same limit (1/6), since d(x-3)/dx = dx/dx = 1.

Answer 2

The lim of x-->3 of (x/(x^2-9)) is infinity.

Answer 3

lim x---> 3
x/(x^2-9) =3(9-9)=3/0=infinity

Answer 4

Can't use l'Hospital unless the limit is 0 for both the top and bottom parts (or infinity for both).

The top part tends towards 3.
The bottom part tends towards 0.

So this goes to infinity.

Answer 5

by L'Hospital's Rule:
dx/dx = 1
d(x^2 - 9)/dx = 2x
lim(1/2x) = 1/6
x-->3

lim[x/x^2 - 9)] = 1/6
x-->3