Limit of ratio with square root in the denominator

I've attempted many methods of solving this limit problem.

But I feel I'm just guessing now, flailing my arms about like a newborn child.

How would you procedurally solve this problem? I'd like to know where to look first in the future. $$\lim_{t\to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}$$

• You want to put the expression over a common denominator, and then multiply by the conjugate of the difference involving a square root. – user84413 Sep 30 '14 at 19:40

$$\frac{1-\sqrt{t+1}}{t\sqrt{t+1}}=\frac{1-(t+1)}{t\sqrt{t+1}(1+\sqrt{t+1})}$$
Cancel $t$ as $t\ne0$ as $t\to0$
$$\lim_{t \to 0} \frac1{t}\left( (1+t)^{-\frac12} -1 \right) =\lim_{t \to 0} \frac1{t}\left( -\frac12 t + \frac38 t^2 +\dots\right) = -\frac12$$