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I'm trying to solve a limit problem but I've never encountered the one like this before.

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

I multiply the right side by $\frac{t\sqrt{1+t}}{t\sqrt{1+t}}$ and combine the terms to get:

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

I can factor out the $t\sqrt{1+t}$ and I am left with:

$$1 - 1 = 0$$

This seems wrong to me. I can't explain why, maybe I just feel that getting a zero after all that work seems lame. Am I following the right steps?

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  • $\begingroup$ Why would you multiply by $\frac{t \sqrt{1+t}}{t \sqrt{1+t}}$ when the right-hand term already has a $t$ in the denominator. Additionally, when you do, you don't get the denominator you claim to get. $\endgroup$ – Eric Towers Jun 16 '14 at 5:14
  • $\begingroup$ @Irresponsible, limit is $t\rightarrow 0$. $\endgroup$ – L.K. Jun 16 '14 at 5:15
  • $\begingroup$ This is in the form $\frac{0}{0}$ $\endgroup$ – Debashish Jun 16 '14 at 5:20
  • $\begingroup$ See also math.stackexchange.com/q/60341 $\endgroup$ – Martin Sleziak Nov 14 '15 at 13:18
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$$\lim_{t\to 0}\left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right)=\lim_{t\to 0}\left[\frac{1}{t}\left(\frac{1}{\sqrt{1+t}}-1\right)\right]=\lim_{t\to 0}\left[\frac{1}{t}\left(\frac{1-\sqrt{1+t}}{\sqrt{1+t}}\right)\right]=\lim_{t\to 0}\left[\frac{1}{t}\left(\frac{1-\sqrt{1+t}}{\sqrt{1+t}}\right)\frac{1+\sqrt{1+t}}{1+\sqrt{1+t}}\right]=\lim_{t\to 0}\left[\frac{1}{t}\left(\frac{1-{1-t}}{\sqrt{1+t}}\right)\frac{1}{1+\sqrt{1+t}}\right]=\lim_{t \to 0}\frac{-1}{\sqrt{1+t}(1+\sqrt{1+t})}=\frac{-1}{2}$$

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  • $\begingroup$ first limit you wrote, will be $t\rightarrow 0$ $\endgroup$ – L.K. Jun 16 '14 at 5:18
  • $\begingroup$ yeah!! Thanks @lavkush $\endgroup$ – tattwamasi amrutam Jun 16 '14 at 5:19
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Using L'Hopital's rule,

$$\begin{align} \lim_{t\to 0}\left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right)&=\lim_{t\to 0}\frac{\frac{1}{\sqrt{1+t}}-1}{t}\\ &=\lim_{t\to 0}\frac{\frac{d}{dt}\left(\frac{1}{\sqrt{1+t}}-1\right)}{\frac{d}{dt}\left(t\right)}\\ &=\lim_{t\to 0}\frac{-1}{2(t+1)^{3/2}}\\ &=-\frac12. \end{align}$$

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Apply L' Hospital rule . . The answer is $- \frac{1}{2}$

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  • $\begingroup$ This answer was shown earlier, with more explanation, by David H. $\endgroup$ – robjohn Jun 16 '14 at 16:22
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If you learnt Taylor series, the problem can be adddressed in a very simple form since, built around $t=0$, $$\frac{1}{\sqrt{1+t}}=1-\frac{t}{2}+O\left(t^2\right)$$ from which the remaining becomes very simple.

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Once you corrected your second formula to $$ \begin{align} \lim_{t\to0}\left(\frac1{t\sqrt{1+t}}-\frac1t\right) &=\lim_{t\to0}\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}\\ &=\lim_{t\to0}\frac{1-\sqrt{1+t}}{t}\ \lim_{t\to0}\frac1{\sqrt{1+t}} \end{align} $$ You can use L'Hospital on the left limit and just plug $t=0$ into the right limit.

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Let me show you a solution that computes the right limit $t\to 0^+$ This is the my solution. Here I use a trigonometric substitution.

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