Calculate the following limit without L'Hôpital's rule

Question

I need to calculate the following limit:

$$ \lim_{x\to3^+}\frac{\sqrt{x}-\sqrt{3}+\sqrt{x-3}}{\sqrt{x^2-9}} $$

I tried to multiply and divide by $\sqrt{x}-\sqrt{3}-\sqrt{x-3}$ but it didn't help me.

How can I calculate it?

Answer 1

$$ \frac{\sqrt{x}-\sqrt{3}+\sqrt{x-3}}{\sqrt{x^2-9}} =\frac {\sqrt {x-3}} {\sqrt {x+3} (\sqrt x +\sqrt 3)}+\frac 1 {\sqrt {x+3}} \to \frac 1 {\sqrt 6} $$

Answer 2

Note that $\sqrt{x^2-9}=\sqrt{x+3}\sqrt{x-3}$ and that therefore\begin{align}\lim_{x\to3^+}\frac{\sqrt x-\sqrt3+\sqrt{x-3}}{\sqrt{x^2-9}}&=\lim_{x\to3^+}\left(\frac{\sqrt x-\sqrt3}{\sqrt{x+3}\sqrt{x-3}}+\frac1{\sqrt{x+3}}\right)\\&=\lim_{x\to3^+}\left(\sqrt{\frac{x-3}{x+3}}\frac{\sqrt x-\sqrt3}{x-3}+\frac1{\sqrt{x+3}}\right).\end{align}Now, use the fact that$$\lim_{x\to3^+}\frac{\sqrt x-\sqrt3}{x-3}=\frac1{2\sqrt3},$$since, if you differentiate $\sqrt x$, you get $\frac1{2\sqrt x}$.

Answer 3

Multiplying the numerator and denominator by a conjugate is a good approach, but the multiplier should have a nonzero limit in order to avoid introducing additional zero factors. Here your chosen multiplier $\sqrt x-\sqrt3-\sqrt{x-3}$ fails that requirement.

Instead rearrange the numerator to read $\sqrt x+\sqrt{x-3}-\sqrt3$, and conjugate the nonzero $\sqrt3$ term so your multiplier is $\sqrt x+\sqrt{x-3}+\sqrt3$. This leads to

$\dfrac{\sqrt x+\sqrt{x-3}-\sqrt3}{\sqrt{x^2-9}}=\dfrac{(\sqrt x+\sqrt{x-3})^2-3}{\sqrt{x^2-9}(\sqrt x+\sqrt{x-3}+\sqrt3)}=\dfrac{2x-6+2\sqrt x\sqrt{x-3}}{\sqrt{x^2-9}(\sqrt x+\sqrt{x-3}+\sqrt3)}$

Then with $x^2-9=(x+3)(x-3)$ and $2x-6=2(x-3)$, we cancel out a factor of $\sqrt{x-3}$ leaving a denominator that no longer $\to0$ as $x\to3$. We then substitute accordingly:

$\dfrac{2\sqrt{x-3}+2\sqrt x}{\sqrt{x+3}(\sqrt x+\sqrt{x-3}+\sqrt3)}\to\dfrac{2\sqrt3}{\sqrt6(2\sqrt3)}=\color{blue}{\dfrac1{\sqrt6}}.$