Cubic Function and Epsilon-Delta Definition

Question

I’ve tried adding terms and construct a perfect square but failed. I’m stuck.

How can I show that this limit equals four by using epsilon-delta definition.

$$\lim_{x\to 1} (x^3 + 4x^2 - 1) = 4$$

Answer 1

Hint: $$ (x^3 + 4x^2 - 1)-4 = x^3 + 4x^2 - 5 = (x-1)(x^2 + 5) $$ The first factor can be made smaller than any $\delta$ you please. What about the second?

Answer 2

It can be easier to manage by shifting the argument. Let $x=y+1$, then

$$(x^3+4x^2-1)-4=y(y^2+7y-5), $$ and you take the limit for $y\to0$.

Then for $|y|<1$, $|y^2+7y-5|<11$ (because the function in the absolute value is growing, and has the minimum $-11$ at $x=-1$).

So in this range,

$$|y(y^2+7y-5)|<|11y|.$$