Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

I attemped to combine the two fraction and multiply by the conjugate and I ended up with:


I couldn't really work it out in my head on what to do with the last term $t\sqrt{1+t}({t\sqrt{1+t}})$ so I left it like that because I think it works anyways. Everything is mathematically correct up to this point but does not give the answer the book wants yet. What did I do wrong?

share|cite|improve this question
As $x$ approaches $0$ ?? $x=t$, eh? – GEdgar Jan 18 '12 at 1:16
Something has gone wrong with your algebra. Can you list out the steps you took in more detail? – Joe Johnson 126 Jan 18 '12 at 1:22
up vote 8 down vote accepted

Perhaps you were trying something like

$\dfrac{1}{t\sqrt{1+t}} - \dfrac{1}{t} = \dfrac{1-\sqrt{1+t}}{t\sqrt{1+t}} = \dfrac{1-(1+t)}{t\sqrt{1+t}(1+\sqrt{1+t})} = \dfrac{-1}{\sqrt{1+t}(1+\sqrt{1+t})} $

which has a limit of $\dfrac{-1}{1 \times (1+1)} = -\dfrac{1}{2}$ as $t$ tends to $0$.

Added: If you are unhappy with the first step, try instead $\dfrac{1}{t\sqrt{1+t}} - \dfrac{1}{t} = \dfrac{t-t\sqrt{1+t}}{t^2\sqrt{1+t}} = \dfrac{t^2-t^2(1+t)}{t^3\sqrt{1+t}(1+\sqrt{1+t})} = \dfrac{-t^3}{t^3\sqrt{1+t}(1+\sqrt{1+t})} $ $= \dfrac{-1}{\sqrt{1+t}(1+\sqrt{1+t})}$ to get the same result

share|cite|improve this answer
I think you did that wrong, for the fractions to be combined you have to multiply them by each others denominators. – user138246 Jan 18 '12 at 1:16
@Jordan: The common denominator is $t\sqrt{1+t}$. You can do it, as you say, to get $t^2\sqrt{1+t}$. You'll just have an extra factor of $t$ in the numerator. – Joe Johnson 126 Jan 18 '12 at 1:20
@Jordan Henry used a least common denominator:$${1\over t\sqrt{1+t}}-{1\over t}={1\over t\sqrt{1+t}}-{\sqrt{1+t}\over t\sqrt{1+t} } = { 1-\sqrt{1+t}\over t\sqrt{1+t}}$$ – David Mitra Jan 18 '12 at 1:21
I am not really following what is happening or how that is a valid operation. The rule I have always heard is that you have to multiply be both the denominators or a lcd which is logical to me. If I have 1/2 + 1/4 I can make it 2/4 + 1/4 which works out. – user138246 Jan 18 '12 at 1:24
@Jordan you can multiply by what is necessary to get both denominators the same. e.g., $$ {1\over 2}+{1\over4}={2\cdot1\over2\cdot 2}+{1\over4 }$$ or $${3\over 6}+ {1\over 15}= {5\cdot 3\over5\cdot6}+{2\cdot1\over 2\cdot15} $$ – David Mitra Jan 18 '12 at 1:35


$$\begin{align} \frac{1}{\sqrt{1+t}} &= (1+t)^{-1/2} = 1 - \frac{1}{2}\;t + o(t) \\ \frac{1}{t\sqrt{1+t}} &= \frac{1}{t} - \frac{1}{2} + o(1) \\ \frac{1}{t\sqrt{1+t}} - \frac{1}{t} &= - \frac{1}{2} + o(1) . \end{align}$$

share|cite|improve this answer
I don't know what that word means or what happened at all here. – user138246 Jan 18 '12 at 1:24
The Binomial Theorem says that $(1+t)^{-1/2}=1-\frac12t+o(t)$ where $o$ is little-o. The rest is division and subtraction. – robjohn Jan 18 '12 at 1:46
+1, Been waiting for limit problems to be squashed just like this for a long time, finally the wait is over! – Arjang Jan 18 '12 at 2:06
@Jordan: If you don't know, then ask! – JavaMan Jan 18 '12 at 4:11
The signs $\sim$ should be $=$. – Did Jan 18 '12 at 6:42

I'd use a substitution to get rid of the surd.

$$\mathop {\lim }\limits_{t \to 0} -\frac{1}{t}\left( {1 - \frac{1}{{\sqrt {t + 1} }}} \right) = $$

$$\sqrt {t + 1} = u$$

$$\mathop {\lim }\limits_{u \to 1} -\frac{1}{{{u^2} - 1}}\left( {1 - \frac{1}{u}} \right) = $$

$$\mathop {\lim }\limits_{u \to 1} -\frac{1}{{{u^2} - 1}}\left( {\frac{{u - 1}}{u}} \right) = $$

$$\mathop {\lim }\limits_{u \to 1} -\frac{1}{{u + 1}}\left( {\frac{1}{u}} \right) = -\frac{1}{2}$$

share|cite|improve this answer

You could also use L'Hopitals rule:

First note that

$\frac{1}{t\sqrt{1+t}} - \frac{1}{t} = \frac{1-\sqrt{1+t}}{t\sqrt{1+t}}$

L'Hopitals rule is that if: $f(x)=0$ and $g(x)=0$ then

$\lim_{t\to x} \frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$

with some provisos that I'll ignore here...

In our case

  • $f(t) = 1 - \sqrt{1+t}$

    So $f'(t) = (-1/2)(1+t)^{-1/2}$ and $f'(0)=-1/2$.

  • $g(t) = t\sqrt{1+t}$

    So $g'(t) = \sqrt{1+t} + (t/2)(1+t)^{-1/2}$ and $g'(0)=1$

So finally we get $f'(0)/g'(0) = -1/2$ as the limit we need.

share|cite|improve this answer
If the OP knew derivatives, then one could simply interpret the original limit as $f'(0)$, where $f$ is the function $f(t) = \frac{1}{\sqrt{1+t}}-1$. – JavaMan Jan 18 '12 at 5:27

Let $f:]0,\infty[\to\mathbb{R}$ given by $$f(x)=\frac{1}{\sqrt{x}}.$$ Then $$\frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\frac{f(1+t)-f(1)}{t},$$ so $$\lim_{t\to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\lim_{t\to 0} \frac{f(1+t)-f(1)}{t}=f'(1).$$ Since $$f'(x)=-\dfrac{1}{2}\cdot x^{-\frac{3}{2}}$$ in $]0,\infty[,$ we get $$\lim_{t\to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\left. -\dfrac{1}{2}\cdot t^{-\frac{3}{2}}\right|_1=-\frac{1}{2}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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