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Evaluating this limit
- Thread starter Asghan
- Start date
Bob Brown MSEE
Full Member
- Joined
- Oct 25, 2012
- Messages
- 598
Sup Guys
I`ve been trying to evaluate this limit like crazy, tried cube diff, variable substitution,common factor, rationalizing,etcetc
limx->∞ (1-x3)1/3-x/1-x
Thanks in advance guys
limx->∞ (1-x3)1/3-x/1-x undef x>1
limx->∞ ((1-x3)1/3-x)/(1-x) undef x>1
limx->∞ (1-x3)1/3-x/(1-x) undef x>1
how is problem stated?
Hello, Asghan!
I'll take a guess at what you meant.
Divide numerator and denominator by \(\displaystyle x\!:\)
. . \(\displaystyle \dfrac{\frac{(1-x)^{\frac{1}{3}}}{x} - \frac{x}{x}}{\frac{1}{x} - \frac{x}{x}}\;=\;\dfrac{\frac{(1-x)^{\frac{1}{3}}}{(x^3)^{\frac{1}{3}}} - 1}{\frac{1}{x} - 1} =\;\dfrac{\left(\frac{1-x}{x^3}\right)^{\frac{1}{3}} - 1}{\frac{1}{x} - 1} \;=\;\dfrac{\left(\frac{1}{x^3}-\frac{1}{x^2}\right)^{\frac{1}{3}} - 1}{\frac{1}{x}-1}\)
Therefore: .\(\displaystyle \displaystyle\lim_{x\to\infty} \dfrac{\left(\frac{1}{x^3}-\frac{1}{x^2}\right)^{\frac{1}{3}} - 1}{\frac{1}{x}-1} \;=\;\frac{(0-0)^{\frac{1}{3}}-1}{0-1} \;=\;\frac{-1}{-1} \;=\;1\)
I'll take a guess at what you meant.
\(\displaystyle \displaystyle\lim_{x\to\infty}\frac{(1-x^3)^{\frac{1}{3}} - x}{1-x} \)
Divide numerator and denominator by \(\displaystyle x\!:\)
. . \(\displaystyle \dfrac{\frac{(1-x)^{\frac{1}{3}}}{x} - \frac{x}{x}}{\frac{1}{x} - \frac{x}{x}}\;=\;\dfrac{\frac{(1-x)^{\frac{1}{3}}}{(x^3)^{\frac{1}{3}}} - 1}{\frac{1}{x} - 1} =\;\dfrac{\left(\frac{1-x}{x^3}\right)^{\frac{1}{3}} - 1}{\frac{1}{x} - 1} \;=\;\dfrac{\left(\frac{1}{x^3}-\frac{1}{x^2}\right)^{\frac{1}{3}} - 1}{\frac{1}{x}-1}\)
Therefore: .\(\displaystyle \displaystyle\lim_{x\to\infty} \dfrac{\left(\frac{1}{x^3}-\frac{1}{x^2}\right)^{\frac{1}{3}} - 1}{\frac{1}{x}-1} \;=\;\frac{(0-0)^{\frac{1}{3}}-1}{0-1} \;=\;\frac{-1}{-1} \;=\;1\)