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Problem: $\lim_{t\to0+}{(\frac{1}{t}+\frac{1}{\sqrt{t}})(\sqrt{t+1}-1)}$

I have difficulties to solve this problem. Here are my steps:

$\lim_{t\to0+}{(\frac{1}{t}+\frac{1}{\sqrt{t}})(\sqrt{t+1}-1)}$
=$\lim_{t\to0+}{\frac{(\sqrt{t}+t)(\sqrt{t+1}-1)}{t\sqrt{t}}}$ (To satisfy the condition of l'hôpital Rules)

This is in a $\frac{0}{0}$ form.

$\frac{d}{dy}(\sqrt{t}+t)(\sqrt{t+1}-1)$
=$(\frac{1}{2\sqrt{t}}+1)(\sqrt{t+1}-1)+(\sqrt{t}+t)(\frac{1}{2\sqrt{t+1}})$
$\frac{d}{dy}t\sqrt{t}=\frac{t}{2\sqrt{t}}+\sqrt{t}=\frac{3}{2}\sqrt{t}$

=$\lim_{t\to0+}{\frac{\frac{d}{dy}(\sqrt{t}+t)(\sqrt{t+1}-1)}{\frac{d}{dy}t\sqrt{t}}}=\frac{0}{0}=0$

While the solution gives $\frac{1}{2}$ rather than 0.

I've did the computation again and again but still feel hard to figure out where I made mistakes.

Anyone there to help me? Thanks in advance!!

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    $\begingroup$ Prefer algebraic manipulation first then think of other options. $\endgroup$ – samjoe May 10 '18 at 11:20
  • $\begingroup$ @samjoe thanks, do you mean the problem must in the first step where I attempts to transform the expression into a fraction? $\endgroup$ – ethanlevy97 May 10 '18 at 11:23
  • $\begingroup$ No I mean if $t\to 0$ then $t/t = 1$ and so forth... and rationalise $\sqrt{t+1}-1$ $\endgroup$ – samjoe May 10 '18 at 11:24
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    $\begingroup$ A common denominator for $t$ and $\sqrt t$ is $t$, so $\frac1t+\frac{1}{\sqrt t} = \frac{1+\sqrt t}{t}$. Also, $\sqrt{t+1}-1=\frac{(\sqrt{t+1}-1)(\sqrt{t+1}+1)}{\sqrt{t+1}+1}=\frac{t}{\sqrt{t+1}+1}$. Classic algebraic transformation. $\endgroup$ – Nicolas FRANCOIS May 10 '18 at 11:26
  • $\begingroup$ @Nicolas FRANCOIS: Thanks so much!!!! I found my problem thank you guys! $\endgroup$ – ethanlevy97 May 10 '18 at 11:32
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HINT: $$ \lim_{t\to0+}{\left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)(\sqrt{t+1}-1)} =\lim_{t\to0+}{\left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)\frac{t}{\sqrt{t+1}+1}} $$ Can you see the limit now?

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  • $\begingroup$ Thank you Przemyslaw!! $\endgroup$ – ethanlevy97 May 10 '18 at 11:33
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Observe that $1/\sqrt t$ is neglectible in front of $1/t$ and can be dropped. Then

$$\frac{\sqrt{1+t}-1}{t}=\frac1{\sqrt{1+t}+1}\to\frac12.$$

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Let's transform the expression a bit:

$$\left(\frac 1t+\frac 1{\sqrt t}\right)(\sqrt{t+1}-1) \\ = \frac{1+\sqrt t}t(\sqrt{t+1}-1) \\ = \frac{1+\sqrt t}t\cdot\frac t{\sqrt{t+1}+1} \\ = \frac{1+\sqrt t}{\sqrt{t+1}+1}$$

Now for $t\to0$ we have

$$\frac{1+\sqrt t}{\sqrt{t+1}+1} \to \frac{1+0}{\sqrt{0+1}+1} = \frac 1{1+1}=\frac 12.$$

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