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$\begingroup$

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

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?

  • 1
    Something has gone wrong with your algebra. Can you list out the steps you took in more detail?2012-01-18

5 Answers 5

8

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

  • 0
    @jordan: To add two fractions together, you are correct that you need a common denominator. However, _any_ common denominator will do, and you will save time and energy if you just use the least common denominator. In this case, $t\sqrt{1+t}$ is the _least_ common denominator, so using it will be the quickest and easiest way to subtracting the fractions. Henry did not forget the $t$, it's just that there is already a $t$ in both denominators, so to write both fractions using the LCD, you keep the left fraction the same and multiply the right fraction by $\frac{\sqrt{1+t}}{\sqrt{1+t}}$.2012-01-18
6

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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    Indeed @DidierPiau , Landau gave us a tool to use comfortably $=$ instead of $\sim$.2012-02-02
1

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) = $

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

0

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 $f(t) = \frac{1}{\sqrt{1+t}}-1$.2012-01-18
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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 $\lim_{t\to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\left. -\dfrac{1}{2}\cdot t^{-\frac{3}{2}}\right|_1=-\frac{1}{2}.$