Question

Find the limit of the function algebraically.

limit as x approaches zero of quantity x squared plus three divided by x to the fourth power.

Answer 1

\[\lim_{x \rightarrow 0} \frac{ x^2+3 }{ x^4 }\]

Answer 2

you can take the limit of the numerator, and then limit of the denominator. (I think )

Answer 3

No, you can't do that because the denominator tends to 0.

Answer 4

I mean this,
\[\lim_{x \rightarrow 0 }~\frac{x^2+3}{x^4}\]
\[\Large~\frac{ \lim_{x \rightarrow 0 }~(x^2+3)}{\lim_{x \rightarrow 0 }x^4}\]
\[\Large~\frac{ \lim_{x \rightarrow 0 }~(x^2+3)}{(\lim_{x \rightarrow 0 }x)^4}\]

Answer 5

and it would be equivalent to ∞

Answer 6

You can separate them, but it doesn't help solve the problem.

Answer 7

yes, it doesn't.

Answer 8

I don't find anything better though.

Answer 9

Well, normally what you would do is factor the top in bottom and then cancel out any repeated roots. If the number approached is still a root of the denominator, then you have an asymptote. This means the limit will be undefined, infinite, or negatively infinite.

Answer 10

yes, it is ∞, because it is unfactorable...

Answer 11

(In this case ∞)

Answer 12

In this case we can use: \[
\lim_{x\to 0}\frac 1{|x|} = \infty
\]If we really want to use limit rules.

Answer 13

\[
\lim_{x\to 0}\frac{x^2+3}{x^4} = \lim_{x\to0}\frac{1}{x^2} +3 \lim_{x\to0}\frac{1}{x^4} =
\left(\lim_{x\to0}\frac{1}{|x|} \right)^2+3 \left(\lim_{x\to0}\frac{1}{|x|}\right)^4 = \infty
\]

Answer 14

Though even doing that is a bit sketchy due to the fact that the limits don't approach a finite number.