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.

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7 Answers7


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



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.

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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,


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.


On the 4th row of the second calculation, you arrive at the following expression: $$ \frac{2-x}{2-x} $$ This expression is actually $\frac{0}{0}$ (undefined) at $x=2$. Equating it to $1$ - which is what you did is the same as removing the singularity (the point of where the function is not defined) in the expression and that's why you get the correct limit. It is correct since we have replaced a function that we could not evaluate at $x=2$ with another function that can be evaluated and have the same limit as the one we could not evaluate.

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If you replace with $t:= 2-x$, the question becomes

$$\lim _{t\to 0}\frac{\sqrt{t+4}-2}{\sqrt{t+1}-1}$$

From wolframalpha, reading the diagram for $\sqrt{t+1}-1$, at the neighbour of $0$ it's something like $t$, or $$\sqrt{t+1}-1=t+O(t^2)$$; similarly, $$\sqrt{t+4}-2=\frac12 t+O(t^2)$$ , here we are using the big O notation.

Then $$\lim _{t\to 0}\frac{\sqrt{t+4}-2}{\sqrt{t+1}-1} = \lim _{t\to 0}\frac{\frac12 t+O(t^2)}{t+O(t^2)}=\frac12$$

Notice the last step is reduction of a fraction $t$.

In your first result,

$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indetermination $\frac{0}{0}$

It's actually

$$\lim _{t\to 0} \left(\dfrac{(\sqrt{t+4}-2)(\sqrt{t+1}+1)}{t}\right) $$

, as the reduction hasn't been done, it's still $\frac{0}{0}$.

So the trick is to derive something as $t$ and reduce.

Sometimes, the problem is a bit tricky that need to do more than one round.

ps. in analysis, after limit, derivative would be explained, then its Taylor's expansion, at that time $\sqrt{t+1}-1=\frac12 t+O(t^2)$ will be more obvious.

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  • I will keep those words, thank you @athos – 欲しい未来 Oct 09 '20 at 20:25
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    @欲しい未来: I support this answer's advice that you need to learn asymptotic expansion to really understand what is going on. For some details read [this explanation](https://math.stackexchange.com/q/3585298/21820) carefully, and then go through [this example](https://math.stackexchange.com/a/2504179/21820), and then go through in the examples in all the linked posts. The key always is to get a proper grasp of what is (eventually) big/small, which will tell you what is the limiting behaviour. – user21820 Feb 24 '21 at 08:00
  • But athos, your last sentence is wrong... $\sqrt{1+t} ∈ 1+\color{red}{\frac12}t+O(t^2)$... – user21820 Feb 24 '21 at 08:03
  • @user21820 indeed. thanks for the tip, i've updated the answer. – athos Feb 24 '21 at 18:21

In this case, the rationalisation of the denominator not only simplifies the expression a bit, but reveals a rate of change, which makes the determination of the limit quite fast: $$\frac{(\sqrt{6-x}-2)\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}}{2-x}=\underbrace{\frac{\sqrt{6-x}-2}{2-x}}_{\textstyle\downarrow\atop \textstyle(-\sqrt{6-x})_{x=2}'}\cdot(\underbrace{\sqrt{3-x}+1}_{\textstyle\downarrow\atop\textstyle2}), $$ hence the limit is $\;-\biggl(\dfrac{1}{2\sqrt{6-x}}\biggr)_{\!x=2}\cdot 2=-\dfrac12.$

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If you have an expression $$\frac{f(x)}{g(x)}$$where both $\lim_{x\to a}f(x)=0$ and $\lim{x\to a}g(x)=0$, you get a limit that looks like $\frac 00$. Now in this case, you write $$f(x)=(x-a)f_1(x)\\g(x)=(x-a)g_1(x)$$ Then $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim\frac{(x-a)f_1(x)}{(x-a)g_1(x)}=\lim_{x\to a}\frac{f_1(x)}{g_1(x)}$$ We can simplify the $x-a$ terms, because the expression is non zero, no matter how close we are, except at $x=a$. If your new limit has a solution, then you are done.

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$z:=\sqrt{3-x}$; and consider $\lim z \rightarrow 1;$




Take the limit.

Peter Szilas
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