Limit, t->infty, of (t) [ (1 + 1/t)^(1/3) - 1]

[Jenny4]

Hey, I'm having trouble with this. Any help will be appreciated greatly.
Find

\(\displaystyle \lim_{t \to \infty} t\left(\left(1 + \frac{1}{t}\right)^{\frac{1}{3}} - 1\right)\)

My attempt so far:

\(\displaystyle t\left(\left(1 + \frac{1}{t}\right)^{\frac{1}{3}} - 1 \right) = \left((t^3 + t^2)^{\frac{1}{3}} - t \right)\)

Multiplying by conjugate : \(\displaystyle \left((t^3 + t^2)^{\frac{1}{3}} - t \right) \left((t^3 + t^2)^{\frac{1}{3}} + t \right)\)

\(\displaystyle \frac{(t^3 + t^2)^{\frac{2}{3}} - t^2}{(t^3 + t^2)^{\frac{1}{3}} +t}\)

I have no idea how to find the limit of that, or where to go from here (or indeed if this is the right way to go..)

Many thanks,
Jenny.

 

[royhaas]

Re: Limits

Let \(\displaystyle x=(1+1/t)^{1/3}\). Now multiply and divide by \(\displaystyle x^2+x+1\).

 

[soroban]

Re: Limits

Hello, Jenny!

\(\displaystyle \text{Find: }\;\lim_{t \to \infty} t\left[\left(1 + \frac{1}{t}\right)^{\frac{1}{3}} - 1\right]\)

My attempt so far:

\(\displaystyle t\left[\left(1 + \frac{1}{t}\right)^{\frac{1}{3}} - 1 \right] \;= \;\left(t^3 + t^2\right)^{\frac{1}{3}} - t\) . . . .Good!

Your conjugate was incorrect . . .

There is a cube root in there.
We must use the form: .\(\displaystyle (a - b)(a^2+ab + b^2) \;=\;a^3-b^3\)


Multiply top and bottom by: .\(\displaystyle (t^3+t^2)^{\frac{2}{3}} + t(t^3+t^2)^{\frac{1}{3}} + t^2\)

\(\displaystyle \frac{(t^3+t^2)^{\frac{1}{3}} - t}{1}\cdot\frac{(t^3+t^2)^{\frac{2}{3}} + t(t^3+t^2)^{\frac{1}{3}} + t^2}{(t^3+t^2)^{\frac{2}{3}} + t(t^3+t^2)^{\frac{1}{3}} + t^2} \;=\;\frac{(t^3+t^2) - t^3}{(t^3+t^2)^{\frac{2}{3}} + t(t^3+t^2)^{\frac{1}{3}} + t^2}\)

. . \(\displaystyle = \;\frac{t^2}{(t^3+t^2)^{\frac{2}{3}} + t(t^3+t^2)^{\frac{1}{3}} + t^2} \;=\;\frac{t^2}{\left[t^3\left(1 + \frac{1}{t}\right)\right]^{\frac{2}{3}} + t\left[t^3\left(1 + \frac{1}{t}\right)\right]^{\frac{1}{3}} + t^2}\)

. . \(\displaystyle = \;\frac{t^2}{t^2\left(1+\frac{1}{t}\right)^{\frac{2}{3}} + t^2\left(1 + \frac{1}{t}\right)^{\frac{1}{3}} + t^2} \;=\; \frac{t^2}{t^2\left[(1+\frac{1}{t})^{\frac{2}{3}} + (1 + \frac{1}{t})^{\frac{1}{3}} + 1\right]}\)

. . \(\displaystyle = \;\frac{1}{(1+\frac{1}{t})^{\frac{2}{3}} + (1 + \frac{1}{t})^{\frac{1}{3}} + 1}\)


\(\displaystyle \text{Then: }\;\lim_{t\to\infty}\,\frac{1}{(1+\frac{1}{t})^{\frac{2}{3}} + (1 + \frac{1}{t})^{\frac{1}{3}} + 1} \;=\;\frac{1}{(1+0)^{\frac{2}{3}} + (1+0)^{\frac{1}{3}} + 1} \;=\;\frac{1}{1+1+1} \;=\;\boxed{\frac{1}{3}}\)

 

[Jenny4]

Re: Limits

Thank you both so much. x

I don't think I could've spotted that.