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.

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.

anonymous · Answer

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

solomonzelman · Answer

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

anonymous · Answer

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

solomonzelman · Answer

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

solomonzelman · Answer

and it would be equivalent to ∞

anonymous · Answer

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

solomonzelman · Answer

yes, it doesn't.

solomonzelman · Answer

I don't find anything better though.

anonymous · Answer

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.

solomonzelman · Answer

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

solomonzelman · Answer

(In this case ∞)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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