# Limits lim x→3+ = 81-x4/(x2-6x+9)2

## Homework Statement

Calculate the limit, if it exists:

lim x→3+ = 81-x4/(x2-6x+9)2

## The Attempt at a Solution

-(x4-81)/(x-3)(x-3)

= -(x2-9)(x2+9)/(x-3)(x-3)

= -(x-3)(x+3)(x2+9)/(x-3)(x-3)

= -(x+3)(x2+9)/(x-3)..............I'm stuck here and don't know what to do next

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You've already done the first step: that is to cancel out common factors in the numerator and denominator so that you don't get 0/0.

HallsofIvy
Homework Helper
You have, as Vahsek said, already cancelled what you could so that you no longer have "0/0" when you set x= 3. What do you get? Remember that the problem was to "Calculate the limit, if it exists".

By substituting 3 in the denominator, it would be undefined, so their is no limit ?

By substituting 3 in the denominator, it would be undefined, so their is no limit ?
Indeed, substituting 3 means that you would have 0 in the denominator, so the original expression would diverge as x → 3+. Therefore, you could argue that there is no limit since the expression would not converge to any real number per say.

That being said, I feel that a more precise answer is needed. For instance, you could say how it diverges.

Indeed, substituting 3 means that you would have 0 in the denominator, so the original expression would diverge as x → 3+. Therefore, you could argue that there is no limit since the expression would not converge to any real number per say.

That being said, I feel that a more precise answer is needed. For instance, you could say how it diverges.
I'm guessing it would go to negative infinity then ?

I'm guessing it would go to negative infinity then ?
That's right!

Ok, thanks guys for your help :)