# Finding limit of two quotients

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

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

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You have a couple typos, so it's not clear what you intend. –  Bill Dubuque Aug 28 '11 at 15:38
6 questions in 12 hours. Don't you think that it might be more beneficial if you tried at least some of your homework yourself? I'm sure that you are not expected to solve each of the questions within five minutes without thinking. Why don't you spend a day or two trying? –  Alex B. Aug 28 '11 at 15:43
If you don't learn to do any of these problems yourself, then you are wasting your time in that class anyway. The purpose of the class is not to become efficient at using stackexchange, google, and co. –  Alex B. Aug 28 '11 at 15:51
Your prior five questions were all limits of a quotient (fraction). Do you know how to transform the above difference of fractions into a single fraction? –  Bill Dubuque Aug 28 '11 at 16:02
@Jordan Carlyon:I guess your problem is not calculus or limit your problem is in algebra-precalculus.You may try revising it once before moving onto further calculus. –  VelvetThunder Aug 28 '11 at 16:30

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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@Jordan Andre has used two common techniques here: finding a common denominator (to get the second equation) and rationalizing (to get the fourth equation). These are the kinds of algebraic "tricks" that you want to try to pick up during a Calculus course. You've used the word "memorize" quite a bit, but it must be more like "intuition". Take from this example (and others like it) the rules of thumb that finding common denominators and rationalizing are good things to try when evaluating limits, and soon you will find yourself using these techniques without thinking about it. –  Austin Mohr Aug 28 '11 at 16:12
I don't think this is something I can learn, I have tried for a long time and never commited it to memory. –  user138246 Aug 28 '11 at 16:16
@André When there is an ongoing Socratic dialogue in the comments - as above - it would be best to refrain from giving a complete answer while this is still active. This destroys the potential learning experience and wastes the time of those who participated in such. –  Bill Dubuque Aug 28 '11 at 16:34
@Jordan There is no sense in taking the stance that you can't learn something. It sounds as though you just need a little brushing up on some concepts from algebra. Is there a tutoring center or something like that at your school? Such a thing would allow you to solve many more problems per hour than stackexchange. –  Austin Mohr Aug 28 '11 at 16:41
@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. –  Bill Dubuque Aug 28 '11 at 17:01

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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That won't help since, per his prior questions, he hasn't yet learned derivatives. See the many comments there and here. –  Bill Dubuque Aug 30 '11 at 3:24
@Bill: After all your comments viewing limits as derivatives, I can't help but think of them in the same light. –  JavaMan Aug 30 '11 at 3:44
Conjugating the numerator is in essence converting it to a limit that's a derivative (of a polynomial), times a cofactor with finite limit. E.g. from his prior question $$\rm \lim_{x\:\to\: 0}\ \dfrac{\sqrt{f(x)}-\sqrt{g(x)}}{x}\ = \ \lim_{x\:\to\: 0}\ \dfrac{f(x)-g(x)}{x\ (\sqrt{f(x)}+\sqrt{g(x)})}\ =\ \dfrac{f_1-g_1}{\sqrt{f_0}+\sqrt{g_0}}$$ That's just $\rm\:\lim_{x\to0}(f-g)/x= (f-g)'(0)\:$ over $\rm\:\sqrt{f_0}+\sqrt{g_0}\:.\:$ But this last limit may be computed without knowledge of derivatives by cancelling $\rm\:x\:$ and evaluating the (polynomial) result at $\rm\:x=0\:.\quad$ –  Bill Dubuque Aug 30 '11 at 4:31
Here's another example, of rationalizing the numerator, namely specializing the quadratic formula when $\: a = 0\:.\:$ –  Bill Dubuque Aug 30 '11 at 20:40

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 $\to 1$.

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