Limit of $\frac{(\sqrt{x^2+12}-4)}{2-\sqrt{x^3-4}}$ as $x$ goes to 2

Question

$$\lim_{x\to 2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$$

I know it must be very trivial, but yet I'm stuck at this problem and other similar problems for quite a long time. I'd be glad if someone showed me a full solution without using L'Hospital and delighted if someone could give some hints about how to atack this kind of problem... what is the thinking behind? Is it just about intuition? Or is it about test all sort of algebraic manipulation?

Answer 1

Try that $$\lim_{x \to 2}\frac{\sqrt{x^2+12}-4}{2 - \sqrt{x^3-4}} = \lim_{x \to 2 }\frac{(\sqrt{x^2+12}-4)(\sqrt{x^2+12}+4)(2 + \sqrt{x^3-4})}{(2 - \sqrt{x^3-4})(\sqrt{x^2+12}+4)(2 + \sqrt{x^3-4})} = \lim_{x \to 2}\frac{(x^2-4)(2 + \sqrt{x^3-4})}{(8-x^3)(\sqrt{x^2+12}+4)}$$

Then you can use that the limit of a product is a product of the limits.

$$=\underbrace{\lim_{x\to 2}\frac{2+\sqrt{x^3-4}}{(4 + \sqrt{x^2+12})}}_{=1/2}\lim_{x \to 2}\frac{(x^2-4)}{(8-x^3)} = -\frac{1}{2}\underbrace{\lim_{x \to 2}\frac{x+2}{x^2+2x+4}}_{=1/3} = -\frac{1}{6}$$

Where, after using $8-x^3 = - (x^3-8)$ and $x^2-4=(x-2)(x+2)$ we noted that

$$(x^3 - 8) = (x - 2)(x^2 + 2x + 4)$$

In general you are correct in the way of treat this kind of problems. This type of manipulations of polynomials are very useful: when you get a polynomial such as $x^3-8$ and it is easy to see that $x=2$ is a root you can divide by $x-2$ and lower the order of the polynomial.

Answer 2

Since you already received good answers, let me show you another way.

Since $x\to 2$, define $x=y+2$ which makes $$A=\lim_{x\to 2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}=\lim_{y\to 0}\frac{\sqrt{y^2+4 y+16}-4}{2-\sqrt{y^3+6 y^2+12 y+4}}$$ Since $y$ is small, we can neglect $y^2$ and $y^3$ which makes $$A\sim\frac{\sqrt{4 y+16}-4}{2-\sqrt{12 y+4}}$$ Now, using the generalized binomial theorem (or Taylor series) $$\sqrt{4 y+16}=4+\frac{y}{2}+\cdots$$ $$\sqrt{12 y+4}=2+3 y+\cdots$$ from which you can easily conclude.

Answer 3

You need to get used to the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which solves these algebraic limit problems instantly.

Thus we have \begin{align} L &= \lim_{x \to 2}\frac{\sqrt{x^{2} + 12} - 4}{2 - \sqrt{x^{3} - 4}}\notag\\ &= \lim_{x \to 2}\frac{\sqrt{x^{2} + 12} - \sqrt{16}}{(x^{2} + 12) - 16}\cdot\frac{x^{2} - 4}{8 - x^{3}}\cdot\frac{4 - (x^{3} - 4)}{\sqrt{4} - \sqrt{x^{3} - 4}}\notag\\ &= -\lim_{t \to 16}\frac{t^{1/2} - 16^{1/2}}{t - 16}\cdot\lim_{x \to 2}\frac{x^{2} - 2^{2}}{x - 2}\cdot\frac{x - 2}{x^{3} - 2^{3}}\cdot\lim_{u \to 4}\frac{u - 4}{u^{1/2} - 4^{1/2}}\text{ (}u = x^{3} - 4, t = x^{2} + 12)\notag\\ &= -\frac{1}{2}\cdot 16^{-1/2}\cdot 2\cdot 2^{1}\cdot\frac{1}{3\cdot 2^{2}}\cdot\dfrac{1}{\dfrac{1}{2}\cdot 4^{-1/2}}\notag\\ &= -\frac{1}{3}\cdot\frac{2}{4}\notag\\ &= -\frac{1}{6}\notag \end{align}

Answer 4

Let $f(x) = \sqrt { x^2 + 12}, g(x) = \sqrt { x^3 -4}.$ The expression equals

$$\tag 1-\frac{f(x) - f(2)}{g(x) - g(2)} = -\,\,\dfrac{\dfrac{f(x) - f(2)}{x-2}}{\dfrac{g(x) - g(2)}{x-2}}.$$

By definition of the derivative, the numerator on the right of $(1)$ has limit $f'(2),$ the denominator has limit $g'(2).$ Thus the desired limit is $-f'(2)/g'(2),$ an easy computation.