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$\lim_{t \to 0} \left(\frac{1}{t\sqrt{1+t}}- \frac 1 t \right)$

I am attempting to find the limit but I am not sure where to even start, I have tried previous methods but they do not seem to help.

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    See also http://math.stackexchange.com/q/8356662015-11-14

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I assume that you want to find $\lim_{t\to 0}\left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right).$ Bring the expression we are interested in to the common denominator $t\sqrt{1+t}$. So we want to find $\lim_{t\to 0}\left(\frac{1}{t\sqrt{1+t}}-\frac{\sqrt{1+t}}{t\sqrt{1+t}}\right),$ that is, $\lim_{t\to 0}\left(\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}\right).$ This has a familiar look! Multiply "top" and "bottom" by $1+\sqrt{1+t}$. So we want to find $\lim_{t\to 0}\frac{(1-\sqrt{1+t})(1+\sqrt{1+t})}{(t\sqrt{1+t})(1+\sqrt{1+t})}.$ The top simplifies to $-t$. If $t\ne 0$, we can then cancel $t$ from top and bottom. We are trying to find $\lim_{t\to 0}\frac{-1}{(\sqrt{1+t})(1+\sqrt{1+t})}.$ Now we can safely let $t$ approach $0$: $\lim_{t\to 0}\frac{-1}{(\sqrt{1+t})(1+\sqrt{1+t})}=-\frac{1}{2}.$

Another way: (sketch) Rewrite our expression as $\frac{\dfrac{1}{\sqrt{1+t}}-1}{t}.$ Multiply top and bottom by $\dfrac{1}{\sqrt{1+t}}+1$, and simplify. We get cancelling $t$'s like before.

Comment: The two expressions $1/(t\sqrt{1+t})$ and $1/t$ that we start with each behave badly as $t$ approaches $0$. Maybe if they are brought together, their badness will cancel out. That is one reason for bringing to a common denominator. Another reason is the familiar one from arithmetic. If we want information about the difference between two fractions, we bring them to a common denominator.

Another reason for the common denominator is that as you contemplate the idea in your mind's eye, you can see that something you have dealt with before will end up on top. The more problems you do, the more you will notice familiar patterns when looking at a new problem. Much of this stuff, and particularly the differentiation that will follow, will soon seem pretty mechanical to you.

Calculate: You have not had time to develop intuition about limits. To develop a bit of intuition, use a calculator to evaluate our function for $t$ nearish to $0$. Don't pick too tiny a $t$, like $t=10^{-9}$, because roundoff error in the calculator will give you misleading results. Pick instead something like $t=0.01$, or $t=-0.005$, or $t=10^{-4}$. Your function evaluations should give answers close to $-1/2$. That's one way to check, by calculator, whether one has made a mistake in the evaluation of the limit.

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    @André Sometimes such dialogues can prove very useful to identify gaps in knowledge, flawed reasoning, etc. But once a complete answer is posted that process becomes much more difficult. Based on the OP's prior questions, I thought it was high time to attempt to get to the root of the his problems.2011-08-28
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HINT:

\lim_{t \to 0} \left( \frac{1}{t \sqrt{1 + t}} - \frac{1}{t} \right) = \lim_{t \to 0} \frac{\frac{1}{\sqrt{1 + t}} - 1}{t} = f'(0)

where $f(x) = \frac{1}{\sqrt{1+x}}$.

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    Here's [another example,](http://math.stackexchange.com/questions/60792/x-frac-b-pm-sqrtb2-4ac2a-show-that-x-c-b-when-a-0/60799#60799) of **rationalizing the numerator**, namely specializing the quadratic formula when $\: a = 0\:.\:$2011-08-30
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HINT $\ $ Per comments, over a common denominator it is $\rm\: \dfrac{1-\sqrt{1+t}}t\: \dfrac{1}{\sqrt{1+t}}$

By your prior problem, as $\rm\:t\to 0\:$ the first factor $\rm\to -1/2\:.\:$ The second factor $\to 1$.