# Finding the limit of $\frac{1}{t\sqrt{1+t}} - \frac{1}{t}$ as $t$ tends to $0$

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

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

$$\frac{t^2-t^2\sqrt{1+t}}{t^3+{t\sqrt{1+t}({t\sqrt1+t})}}$$

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?

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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

## 5 Answers

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

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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

Asymptotics:

\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}

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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: en.wikipedia.org/wiki/Asymptotic_analysis. 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.

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

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

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

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

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

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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.

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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 $$\lim_{t\to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\left. -\dfrac{1}{2}\cdot t^{-\frac{3}{2}}\right|_1=-\frac{1}{2}.$$

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