2
$\begingroup$

I have been at this one for a while and I can't make it work. I am trying to evaluate the limit of $(\sqrt{1+t}-\sqrt{1-t})/t$ as t approaches 0. It seems pretty simple but I can't make it work. I have tried a lot of things but I think what I am meant to do is use the conjugate $(\sqrt{1+t}+\sqrt{1-t})$ which gives me $(1+t)-(1-t)/t(\sqrt{1+t}+\sqrt{1-t})$

Which then leaves me with $2t/t(\sqrt{1+t}+\sqrt{1-t})$

$\endgroup$
  • 2
    $\begingroup$ Cancel those $t$'s $\endgroup$ – David Mitra Jan 14 '12 at 23:50
  • $\begingroup$ Then I am left with t on top. $\endgroup$ – toby yeats Jan 14 '12 at 23:51
  • 1
    $\begingroup$ ${2t\over t(\sqrt{1+t}+\sqrt{1-t})}={2\over(\sqrt{1+t}+\sqrt{1-t})}$, for $t\ne0$. $\endgroup$ – David Mitra Jan 14 '12 at 23:52
  • $\begingroup$ I am terrible at math, I thought for a good 45 minutes that $\frac{2t}{t}=t$ since I had 2 ts being divided by one t that a single t would cancel out. $\endgroup$ – toby yeats Jan 14 '12 at 23:56
  • 2
    $\begingroup$ There's really no need to use the word "cancel". It seems to do more harm than good. The thing that's really going on is that multiplication and division are inverse operations, meaning that multiplying and then dividing by the same (nonzero) number results in $1$. So: $\frac{2t}{t} = (2 \cdot t) \div t = 2 \cdot (t \div t) = 2 \cdot 1 = 2$ $\endgroup$ – Austin Mohr Jan 15 '12 at 5:48
1
$\begingroup$

You're on the right track with using the conjugate to get $$\frac{2t}{t(\sqrt{1+t}+\sqrt{1-t})}.$$ As has been said in the comments, this is equal to $$\frac{2}{(\sqrt{1+t}+\sqrt{1-t})}.$$ One way to think about that is $$\frac{2t}{t} = (2 \cdot t) \div t = 2 \cdot (t \div t) = 2 \cdot 1 = 2,$$ as in Austin Mohr's comment. I tend to think about it slightly differently: $$\frac{2t}{t}=\left(2\cdot t\right)\cdot\frac{1}{t}=2\cdot\left(t\cdot\frac{1}{t}\right)=2\cdot1=2$$ (sticking to just multiplication, without division, using the Associative Property of Multiplication, and then using the fact that $t$ and $\frac{1}{t}$ are multiplicative inverses so that their product is $1$). Or, for this particular problem, $$\begin{align} \frac{2t}{t(\sqrt{1+t}+\sqrt{1-t})} &=\left(2\cdot t\right)\cdot\left(\frac{1}{t}\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}}\right) \\ &=2\cdot\left(t\cdot\frac{1}{t}\right)\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}} \\ &=2\cdot 1\cdot\frac{1}{\sqrt{1+t}+\sqrt{1-t}} \\ &=\frac{2}{\sqrt{1+t}+\sqrt{1-t}}. \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.