# Evaluating limits algebraically

I bought the eighth edition of Stewart Calculus (metric version) and I'm up to the section about limits. It's been pretty easy so far, but I've come across a class of limit problems that don't seem solvable with mere algebra. The following is fairly representative of them:

$$\lim_{t\to 0} \frac {\sqrt{1+t}-\sqrt{1-t}}t$$

I tried rationalizing the numerator by multiplying by its conjugate, but I still ended up with a denominator that tends towards 0, and thus I was forced to conclude that the limit did not exist. However, the book kindly gave me the answer of 1, and I can't for the life of me work out how to get to that point via algebraic manipulation.

Have I just misled myself with regards to the quotient limit law? That is, I have been tacitly assuming that the failure of said law amounts to the expression having no limit overall, and now that I type this, it seems like a rather stupid assumption. Does that mean that problems such as these require numerical/graphical methods to solve?

I realize that I might have just answered my own question, but still, I'd like to know if my reflection is accurate. Also, I apologize for the lack of formatting; it's quite late here and as such I found the MathJax instructions... impenetrable.

• You must have made an error in " rationalizing the numerator by multiplying by its conjugate". If you do it correctly the $t$ cancels out. – Michal Adamaszek May 29 at 11:50
• Multilpying by the conjugate, you should get $\lim\limits_{t \rightarrow 0} \dfrac{2t}{t \left( \sqrt{1 + t} + \sqrt{1 - t} \right)} = 1$. – Aniruddha Deshmukh May 29 at 11:50
• Explain your attempt to rationalize the numerator. That method ought to work. – lulu May 29 at 11:51
• @AniruddhaDeshmukh I think it is $1$, not $2$. – Don Thousand May 29 at 11:51
• Hi all, yes I made an elementary mistake - I forgot to account for the fact that a negative times a positive equals a negative, so my numerator ended up being (1+t) + (1-t), which as we all know results in 2, with no t to cancel. This has reinforced how important it is to double check my work, so I don't think it's been a complete waste. Thank you for all your answers :) – Marcus Hendriksen May 29 at 12:44

Rationalizing is a good idea, but it is only a first step: \begin{align}\frac{\left(\sqrt{1+t}-\sqrt{1-t}\right)\color{blue}{\left(\sqrt{1+t}+\sqrt{1-t}\right)}}{t\color{blue}{\left(\sqrt{1+t}+\sqrt{1-t}\right)}} &=\frac{(1+t)-(1-t)}{t\left(\sqrt{1+t}+\sqrt{1-t}\right)} \\[4pt] &=\frac{2t}{t\left(\sqrt{1+t}+\sqrt{1-t}\right)} \end{align} It is normal that you still encounter the same problem when letting $$t\to 0$$ at this point since the only thing we did so far was rewriting, but we didn't actually change anything yet. So you are correct that this denominator still tends to $$0$$ (for $$t \to 0$$).
Now you can simplify by dividing numerator and denominator by the common factor $$\color{red}{t}$$: $$\lim_{t\to 0}\frac{2\color{red}{t}}{\color{red}{t}\left(\sqrt{1+t}+\sqrt{1-t}\right)}=\lim_{t\to 0}\frac{2}{\sqrt{1+t}+\sqrt{1-t}} = \ldots$$ This is probably the crucial step you're missing, unless you made a mistake when rationalizing. Now the denominator no longer tends to $$0$$ for $$t \to 0$$ and you can easily evaluate the limit.
$$\frac{\sqrt{1+t}- \sqrt{1-t}}{t}=\frac{2}{\sqrt{1+t}+ \sqrt{1-t}} \to 1$$ as $$t \to 0.$$