Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$

We are given this limit to evaluate: $$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$

In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this expression by the conjugate of the denominator. $$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .

But multiplying the expression by the conjugate of the demoninator and numerator we get $$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$

$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$

$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$

$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$

$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$

Now we can evaluate the limit: $$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$

Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?

I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.

Thanks in advance for you answer.

Solution 1:

Rationalization is a standard way to manipulate such kind of limits when they lead to an indeterminate form.

The aim for this kind of manipulation is to eliminate the term which leads to the indetermination, indeed by $(A-B)(A+B)=A^2-B^2 \implies A-B= \frac{A^2-B^2}{A+B}$ we have that

$$\sqrt{6-x}-2=\frac{2-x}{\sqrt{6-x}+2}$$

$$\sqrt{3-x}-1=\frac{2-x}{\sqrt{3-x}+1}$$

and by the ratio the problematic $x-2$ term cancel out.

As an alternative we can also use binomial first order approximation (i.e. Taylor's series) at $x=2$ to obtain

$$\sqrt{6-x}=\sqrt{4-(x-2)}=2\sqrt{1-\frac{(x-2)}4}=2\left(1-\frac {x-2}{8}+o(x-2)\right)$$

$$\implies \sqrt{6-x}-2=-\frac {x-2}{4}+o(x-2)$$

$$\sqrt{3-x}=\sqrt{1-(x-2)}=1-\frac {x-2}{2}+o(x-2)$$

$$\implies \sqrt{3-x}-1=-\frac {x-2}{2}+o(x-2)$$

which gives evidence of the same problematic term and the final result.

Solution 2:

As written, no simplification is apparent, and this is due to the presence of the radicals. Now considering the identity

$$a-b=\frac{a^2-b^2}{a+b}$$ there is a hope of getting rid of them. In the case of your numerator,

$$\sqrt{6-x}-2=\frac{(\sqrt{6-x})^2-2^2}{\sqrt{6-x}+2}=\frac{2-x}{\sqrt{6-x}+2}.$$

Now the radical is gone from the numerator and has moved to the denominator, but it is important to notice that it does not cancel because the minus has turned to a plus.

Repeating this trick with the denominator of the original ratio, you will see a simplification.