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.

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    $\begingroup$ You must have made an error in " rationalizing the numerator by multiplying by its conjugate". If you do it correctly the $t$ cancels out. $\endgroup$ – Michal Adamaszek May 29 at 11:50
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    $\begingroup$ Multilpying by the conjugate, you should get $\lim\limits_{t \rightarrow 0} \dfrac{2t}{t \left( \sqrt{1 + t} + \sqrt{1 - t} \right)} = 1$. $\endgroup$ – Aniruddha Deshmukh May 29 at 11:50
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    $\begingroup$ Explain your attempt to rationalize the numerator. That method ought to work. $\endgroup$ – lulu May 29 at 11:51
  • $\begingroup$ @AniruddhaDeshmukh I think it is $1$, not $2$. $\endgroup$ – Don Thousand May 29 at 11:51
  • $\begingroup$ 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 :) $\endgroup$ – Marcus Hendriksen May 29 at 12:44

I tried rationalizing the numerator by multiplying by its conjugate, but I still ended up with a denominator that tends towards 0

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.

  • $\begingroup$ Okay, so it turns out that I'm just an idiot... I ended up with (1+t) + (1-t) for the numerator, which left me with 2... and thus I was not able to factor t from the denominator. I'll pay more attention next time, but thank you for going through this :) $\endgroup$ – Marcus Hendriksen May 29 at 12:42
  • $\begingroup$ No worries and you're welcome! $\endgroup$ – StackTD May 29 at 13:40

$ \frac{\sqrt{1+t}- \sqrt{1-t}}{t}=\frac{2}{\sqrt{1+t}+ \sqrt{1-t}} \to 1$ as $t \to 0.$


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