Help solving a limit in two parts $\lim_{t\to 0}\left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right)$

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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$$\lim_{t\to 0}\left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right)=\lim_{t\to 0}\left[\frac{1}{t}\left(\frac{1}{\sqrt{1+t}}-1\right)\right]=\lim_{t\to 0}\left[\frac{1}{t}\left(\frac{1-\sqrt{1+t}}{\sqrt{1+t}}\right)\right]=\lim_{t\to 0}\left[\frac{1}{t}\left(\frac{1-\sqrt{1+t}}{\sqrt{1+t}}\right)\frac{1+\sqrt{1+t}}{1+\sqrt{1+t}}\right]=\lim_{t\to 0}\left[\frac{1}{t}\left(\frac{1-{1-t}}{\sqrt{1+t}}\right)\frac{1}{1+\sqrt{1+t}}\right]=\lim_{t \to 0}\frac{-1}{\sqrt{1+t}(1+\sqrt{1+t})}=\frac{-1}{2}$$

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}

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

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.

lim┬(t→0)⁡(1/(t√(1+t))-1/t)
Substitute t=tan^2⁡θ. Then your expression becomes,
lim┬(θ→0)⁡〖1/tan^2⁡θ 〗 (1/sec⁡θ -1)
=lim┬(θ→0) ( cos^2⁡θ (cos⁡θ-1))/((1-cos^2⁡θ ) )
=-lim┬(θ→0) cos^2⁡θ/((cos⁡θ+1) )
=-1/2.

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.