# Steps to find limit of $\frac1{t\sqrt{t+1}} - \frac1t$ as $t\to0$

So, it's been about 6 years since I've taken any type of math course, and 7 since I've taken Calculus. Recently, though, I've been trying to relearn Calculus in my free time.

I've been working through my old Calculus textbook, which has been going pretty well. Currently, though, I'm stuck on this problem:

$$\lim_{t\to0} \left(\frac{1}{t\sqrt{t+1}} - \frac1t\right)$$

I need to systematically find the limit as $t\to0$. Right away, I know that I need to manipulate it in such a way that when I plug $0$ in for $t$, I don't have $0$ in the denominator.

I know the answer is $-\frac12$, but I'm having quite a bit of trouble getting there. Could someone help to point me in the right direction? I'm not necessarily asking someone to solve it, I'm just not entirely sure what my first few steps should be.

It's been a while for me, so please keep it in the context of someone who has never worked with limits before.

• Sorry, valverij, I worked a lot yesterday with integrals...here, partial fractions are not appropriate. – Namaste May 3 '13 at 21:36
• You could expand: $$\frac{1}{t}\left(\frac{1}{\sqrt{1+t}}-1\right)=\frac{1}{t}\left(-\frac{t}{2}+ \frac{3t^2}{8}-\cdots \right)$$ – L. F. May 3 '13 at 21:37

$$\frac{1}{t\sqrt{t+1}} - \frac{1}{t} = \frac{1-\sqrt{t+1}}{t\sqrt{t+1}} = \frac{1-\sqrt{t+1}}{t\sqrt{t+1}} \frac{1+\sqrt{t+1}}{1+\sqrt{t+1}}=\frac{-t}{(t\sqrt{t+1})(1+\sqrt{t+1})}=\frac{-1}{(\sqrt{t+1})(1+\sqrt{t+1})}$$
• Thanks, I already see where I when wrong. I kept flipping the $1-\sqrt{t+1}$ around to read $-\sqrt{t+1}+1$ in the numerator in that first/second step, and it was tripping me up down the road. – valverij May 3 '13 at 21:44
If you rewrite the expression (by reducing all to the same denominator), you get $$\Delta(t)\stackrel{\text{def}}{=}\frac{1-\sqrt{t+1}}{t\sqrt{t+1}} = -\frac{1}{\sqrt{t+1}}\frac{\sqrt{t+1}-1}{t}$$ Now, both the numerator and the denumerator go to $0$ with $t$. A way to consider this, though, is to remark that $\frac{1}{\sqrt{t+1}}$ goes nicely to $1$ when $t\to0$; if you had a way to deal with the remaining $\frac{\sqrt{t+1}-1}{t}$, you could compute the limit.
If this rings a bell, you can try to consider the function $g:x\to\sqrt{x+1}$, and its derivative at $0$ (first observing that it is defined and continuous on $[-1,\infty)$, and differentiable on $(-1,\infty)$).