# Limit Evaluation (Conjugate Method)–Further algebraic manipulation?

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

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?

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

• How can the function both be continuous and defined at x=2 but have a removable discontinuity at x=2... Are we essentially postulating that as we rationalize the expression, the discontinuity becomes removed but as the original expression had stood, in its indeterminate form, there was a discontinuity at f(2)? And if so, what did the conjugate multiplication process do to remove the discontinuity–generally and conceptually? – sardinsky Oct 4 '17 at 16:29
• @sardinsky - Let $f(x) = \frac{x^2-1}{x-1}$ and $g(x)=x+1$. For all $x\ne 1, f(x) = g(x)$ but $f(1)$ is undefined and $g(1)=2$. If you graph $y=f(x)$ and $y=g(x)$ the two functions will be identical except at $x=1$. Hence the discontinuity of $y=f(x)$ at $x=1$ can be removed by defining $f(1)=g(1)=2$. That is essentially what $\lim_{x \to 1}f(x) = 2$ is telling you. – steven gregory Oct 4 '17 at 16:44

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

• @MichaelRozenburg can you please elaborate why (2-x) was added to the expression and how this simplifies out to 1/2 – sardinsky Oct 4 '17 at 15:39
• @sardinsky I added something. See now. – Michael Rozenberg Oct 4 '17 at 15:47
• sardinsky --- he multiplied both the numerator and the denominator of the original fraction by the product of the two conjugates (one being the conjugate of the original numerator, the other being the conjugate of the original denominator). The $(2-x)$ factors arise when you perform the two different conjugate multiplications, one in the new numerator and the other in the new denominator. – Dave L. Renfro Oct 4 '17 at 15:50
• @DaveL.Renfro Thank you for the explanation. But why must we multiply by a product of both conjugates? – sardinsky Oct 4 '17 at 15:52
• I played with this a while and didn't see another "precalculus way". Generally speaking, you use the conjugate method to rewrite something so that a square root is eliminated, which you want to do because square root operation does not behave well with respect to addition or subtraction inside of it --- $\sqrt{a+b}$ cannot be rewritten in a useful way in terms of the separate quantities $\sqrt{a}$ and $\sqrt{b}.$ In the limit you had, each of the square roots was critically linked to stuff that is going to zero. – Dave L. Renfro Oct 4 '17 at 15:57

Just another way:

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

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

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