Evaluate the limit $\lim_{t\to0+}{(\frac{1}{t}+\frac{1}{\sqrt{t}})(\sqrt{t+1}-1)}$

Question

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!!

Answer 1

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?

Answer 2

Observe that $1/\sqrt t$ is neglectible in front of $1/t$ and can be dropped. Then

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

Answer 3

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

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