Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$

Question

Calculate the limit

$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$

I tried to factorise and to simplify, but I can't find anything good. $$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$

Answer 1

You should use L'Hôpital's rule.

Answer 2

Substitute $t=\sqrt{x+2}$; if $x\to2$, then $t\to2$, so, after doing $x=t^2-2$, you get \begin{align} \lim_{t\to2}\frac{t(t^2-2)^2-8}{4-(t^2-2)^2} &=\lim_{t\to2}\frac{t^5-4t^3+4t-8}{4t^2-t^4} \\[4px] &=\lim_{t\to2}\frac{t^3(t^2-4)+4(t-2)}{t^2(4-t^2)} \\[4px] &=\lim_{t\to2}\left(-t-\frac{4}{t^2(2+t)}\right) \end{align}

Answer 3

L'Hôpital's rule is not necessary. Here are the steps $$\lim\limits_{x\to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$ $$=\lim\limits_{x\to 2} \frac{8-x^2\sqrt{x+2}}{(x-2)(x+2)}$$ Let $t=\sqrt{x+2}$, then $$\lim\limits_{t\to 2} \frac{8-\left(t^2-2\right)^2 t}{\left(t^2-4\right)t^2}$$ $$=\lim\limits_{t\to 2} \frac{\left(-t^5+4t^3-4t+8\right)}{(t-2)(t+2)t^2}$$ $$=\lim\limits_{t\to 2} \frac{-(t-2)\left(t^4+2t^3+4\right)}{(t-2)(t+2)t^2}$$ $$=\lim\limits_{t\to 2} \frac{-\left(t^4+2t^3+4\right)}{(t+2)t^2}$$ $$=\lim\limits_{t\to 2} \frac{-\left(t^2+2t+\frac{4}{t^2}\right)}{t+2}$$ $$=-\frac{\left(4+4+1\right)}{2+2}$$ $$=-\frac{9}{4}$$

Answer 4

Since the OP does not know L'Hospital's rule (yet), this approach is probably not the best way. When you multiply top and bottom by $x^2\sqrt{x+2}+8$ you get the fraction: $\frac{x^5+2x^4-64}{(4-x^2)(x^2\sqrt{x+2}+8)}$ Both NUM and DENOM have a factor $x-2$ In case of the NUM, you could perform long division to arrive at $x^4+4x^3+8x^2+16x+32$ and in case of the DENOM, it is of course straight forward. Can you finish it from here?

Answer 5

It seems you may not know l'Hôpital's rule. It's also possible without the need for differentiation. Multiply numerator and denominator with the so called conjugate expression of the numerator:

$$\frac{x^2 \sqrt{x+2} - 8}{4-x^2} = \frac{\left( x^2 \sqrt{x+2} - 8 \right)\left( x^2 \sqrt{x+2} + 8 \right)}{\left(4-x^2\right) \left( x^2 \sqrt{x+2} - 8 \right)}$$

Then simplify via $(a-b)(a+b) =a^2-b^2$ in the numerator:

$$\frac{x^4(x+2)-64}{\left(4-x^2\right) \left( x^2 \sqrt{x+2} - 8 \right)}$$

Now you should factorise so you have a common factor $x-2$ in the numerator and denominator. In the denominator that's easy via $4-x^2 = (2-x)(2+x)=-(x-2)(x+2)$. For the nominator, you know that the polynomial divides $(x-2)$ since $x=2$ is a root of the polynomial. You can use long division or Horner's method for this:

$$x^4(x+2)-64 = x^5+2x^4-64=(x-2) (x^4+4 x^3+8 x^2+16 x+32)$$

So back to the limit:

$$\lim_{x \to 2} \frac{x^2 \sqrt{x+2} - 8}{4-x^2} =\lim_{x \to 2} \frac{(x-2) (x^4+4 x^3+8 x^2+16 x+32)}{-(x-2)(x+2) \left( x^2 \sqrt{x+2} + 8 \right)}$$

Cancel the common factor $x-2$ and plug in $x=2$, simplify:

$$\lim_{x \to 2} \frac{x^4+4 x^3+8 x^2+16 x+32}{-(x+2) \left( x^2 \sqrt{x+2} + 8 \right)} = \ldots = -\frac{9}{4}$$

If you have seen l'Hôpital's rule, the other answers give you an alternative.

Answer 6

For the sake of being concise, I have chosen to omit most of the intermediary algebraic simplifications. Letting $u = x+2$, we get

$$\lim_{u \to 4} \frac{{u^{\frac 52} - 4u^{\frac 32} + 4\sqrt{u} - 8}}{4 - u^2 + 4u - 4}$$

By L'hopital, this is equal to $$\lim_{u \to 4} \frac{\frac {5u^2 - 12u+4}{2 \sqrt{u}}}{4 - 2u} = \lim_{u \to 4} \frac{2 - 5u}{4\sqrt{u}} = -\frac{9}{4}$$