Limit of a function

Mar 2010
5
0
Hi :)
Could someone please help me with the general method of finding limits like this one:

x-> infinity of x^(1/3) [ (x+8)^(2/3) - (x+7)^(2/3)]
 
Mar 2010
107
14
\(\displaystyle \lim_{n\rightarrow\infty} {(x+8)}^{2/3} - {(x+7)}^{2/3} \) will give you \(\displaystyle \infty - \infty \), which is an indeterminate of L'Hôpital's rule. You have to algebraically change it up so as \(\displaystyle \lim_{n\rightarrow\infty} \left({(x+8)}^{2/3} - {(x+7)}^{2/3}\right) \) you get \(\displaystyle \frac{0}{0} \) or \(\displaystyle \frac{\infty}{\infty} \). Finally, take the derivative of the function.
 
Oct 2009
4,261
1,836
Hi :)
Could someone please help me with the general method of finding limits like this one:

x-> infinity of x^(1/3) [ (x+8)^(2/3) - (x+7)^(2/3)]


Take note that for \(\displaystyle x>0\,,\,\,x^{4/3}(A+B)^{4/3}=(Ax+Bx)^{4/3}\) , and also that \(\displaystyle A-B=\left(A^{1/3}-B^{1/3}\right)\left(A^{2/3}+A^{1/3}B^{1/3}+B^{1/3}\right)\) :

\(\displaystyle x^{1/3} [ (x+8)^{2/3} - (x+7)^{2/3}]=\) \(\displaystyle x^{1/3}\,\frac{(x+8)^2-(x-7)^2}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}=\) \(\displaystyle x^{1/3}\,\frac{2x+15}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}=\)

\(\displaystyle \frac{2x^{4/3}+15x^{1/3}}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}\,\frac{1/x^{4/3}}{1/x^{4/3}}=\) \(\displaystyle \frac{2+15/x}{\left(1+8/x\right)^{4/3}+\left(1+8/x\right)^{2/3}\left(1+7/x\right)^{2/3}+\left(1+7/x\right)^{4/3}}\) \(\displaystyle \xrightarrow [x\to\infty]{}2/3\)

Tonio
 
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