lim x-->3 of x^2/(x-3); how is this done?

Answer 1

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lim(x->3) x^2/(x-3) => ∞ => no limits exist.

Edit: To the above answer, L'hospital's can be used only for Indeterminate cases i.e. 0/0 etc. this is not one those types since you have a number divided by zero is ∞ or undefined, and not Indeterminate.

Answer 2

L'Hopital's rule only applies when the numerator and denominator both approach either 0 or ±∞.

lim_{x->3} x^2 = 9, so we cannot use L'Hopital here.

With expressions like this, the limit exists if (x - 3) can be cancelled out of the denominator, producing a function that is continuous at x = 3. However, this cannot be done. So lim_{x->3} (x^2)/(x - 3) DNE.

Answer 3

Note that you can substitute x = 3 directly into the expression to get 3^2/(3 - 3) = 9/0, which is neither defined nor indeterminate. Hence, the limit doesn't exist.

You cannot say the limit is infinity or -infinity either. As x --> 3-, x - 3 --> 0-, so x^2/(x - 3) --> -infinity as x --> 3-. On the other hand, x - 3 --> 0+ as x --> 3+, so x^2/(x - 3) --> +infinity as x --> 3+.

I hope this helps!

Answer 4

as x approaches 3, the numerator approaches 3^2 = 9 and the denominator approaches 0. So, the limit is infinite.

Moreover, if you notice, as x approaches 3 from below (i.e. from values less than 3), the amounts are negative, and vice versa.

So, there is no limit.

Answer 5

it is done by using L-hospital rule

we differentiate numerator and denominator and this becomes

lim x-->3 of 2x/1

which is 6.