Limit Evaluation (Conjugate Method)–Further algebraic manipulation?

Question

$$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$

I have tried evaluating the above limit by multiplying either both the conjugate of numerator and the denominator with no avail in exiting the indeterminate form.

i.e. $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}$$ and conversely $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$

To my suspicions of which–either numerator or denominator–conjugate to multiply by I chose $$\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$

This resulted in $$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{3-x-1}$$

Is it indeterminate? What is the reason for multiplying by a specific conjugate in a fraction (denominator or numerator) and the reason for the conjugate being either a) denominator b) numerator c) or both?

Am I simply practicing incorrect algebra by rationalizing the expression to:

$$\frac{(6-x)(3-x)-2\sqrt{3}+2x+2}{x-2}$$

Or am I failing to delve further and manipulate the expression out of the indeterminate form?

Answer 1

The problem with $\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ is that, when you let $x=2$, you get $\dfrac 00$. So we have to assume that $x \ne 2$. This is not necessarily a problem because $\displaystyle \lim_{x \to x_0}f(x)$ does not care about what happens to $f(x)$ at $x=x_0$.

Notice below that a factor of $(2-x)$ appeared in both the numerator and the denominator and was cancelled out. That leaves us with the rational expresson $\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$ which is equal to $\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ for all all $x \ne 2$ and just so happens to be continuous and defined at $x=2$.

So the function $f(x)=\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ has a removable discontinuity at $x=2$. We can remove that discontinuity by defining $\left. f(2)=\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right|_{x=2}=\dfrac 12$

For all $x\ne 2$ we can say

\begin{align} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} &= \left(\frac{\sqrt{6-x}-2}{1} \cdot \frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) \cdot \left(\frac{1}{{\sqrt{3-x}-1}} \cdot \frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\right) \\ &= \frac{6-x-4}{\sqrt{6-x}+2} \cdot \frac{\sqrt{3-x}+1}{3-x-1} \\ &= \frac{2-x}{\sqrt{6-x}+2} \cdot \frac{\sqrt{3-x}+1}{2-x} \\ &= \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2} \\ \end{align}

So $\displaystyle \lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} =\lim_{x \to 2} \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2} = \frac 24 = \frac 12$

Answer 2

By your idea $$\lim_{x\rightarrow2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x\rightarrow2}\frac{(\sqrt{3-x}+1)(6-x-4)}{(\sqrt{6-x}+2)(3-x-1)}$$ $$=\lim_{x\rightarrow2}\frac{(\sqrt{3-x}+1)(2-x)}{(\sqrt{6-x}+2)(2-x)}=\lim_{x\rightarrow2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}=\frac{2}{4}=\frac{1}{2}$$

Answer 3

Just another way:

Let $\sqrt{6-x}-2=h,\sqrt{3-x}-1=k\implies h\to0,k\to0$

and $$6-4-4h-h^2=x=3-1-2k-k^2\implies-h(4+h)=-k(k+2)\implies\dfrac hk=\dfrac{k+2}{h+4}$$

$$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{h\to0,k\to0}\dfrac hk=\lim_{h\to0,k\to0}\dfrac{k+2}{k+4}=?$$