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

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?

Community
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5 Answers5

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

Henry
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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. –  Jan 18 '12 at 01:16
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    @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. – J126 Jan 18 '12 at 01:20
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    @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 01:21
  • 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. –  Jan 18 '12 at 01:24
  • @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 01:35
  • I still don't see what is going on, it looks like you forgot about the t in the last step and it sort of just goes away when it should be making the denominator 0. –  Jan 18 '12 at 01:40
  • @Jordan: The numerator is $1-(1+t)$ which is $-t$ and so you can cancel the $t$ in the numerator and denominator (or the $t^3$ in the added version) – Henry Jan 18 '12 at 01:42
  • @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 $\frac{\sqrt{1+t}}{\sqrt{1+t}}$. – JavaMan Jan 18 '12 at 04:09
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Asymptotics:

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

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

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

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

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

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

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

Pedro
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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.

Michael Anderson
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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 05:27
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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 $$\lim_{t\to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}=\left. -\dfrac{1}{2}\cdot t^{-\frac{3}{2}}\right|_1=-\frac{1}{2}.$$

leo
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