Need help..urgent! Limit of cubic root..

[nobody888]

Hye.. pls help to solve this question..

The answer is 1/3 but didn't know the solving method..

 

[nobody888]

[FONT=MathJax_Math-italic]f[FONT=MathJax_Main]′[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]lim[/FONT][FONT=MathJax_Math-italic]h[/FONT][FONT=MathJax_Main]→[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]h[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]f[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math-italic]h[/FONT][/FONT]


[FONT=MathJax_Math-italic]f[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]3[/FONT][/FONT] and the limit would evaluate as f'(8)..[FONT=MathJax_Math-italic]f[FONT=MathJax_Main]′[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main])[/FONT][/FONT]

 

[HallsofIvy]

Hye.. pls help to solve this question..

The answer is 1/3 but didn't know the solving method..

HOW you would do this depends upon what you already know. Since this is posted under "Caculus" you may well know that the derivative of function f(x) at x= 8 is defined as \(\displaystyle \lim_{h\to 0}\frac{f(8+h)- f(8)}{h}\) and that the derivative of \(\displaystyle f(x)= x^n\) is \(\displaystyle nx^{n-1}\) so that, in particular, the derivative of \(\displaystyle f(x)= x^{2/3}\) is \(\displaystyle (2/3)x^{-1/3}\). Putting those together makes this problem easy.

But this may also be an exercise preparing you for that. If so this is a much harder algebra problem. You will need to use the fact that \(\displaystyle a^3- b^3= (a- b)(a^2+ ab+ b^2)\) with \(\displaystyle a= (8+ h)^{2/3}\) and \(\displaystyle b= 8^{2/3}\) so that \(\displaystyle (8+h)^2- 8^2= ((8+ h)^{2/3}- 8^{2/3})((8+ h)^{4/3}+ 8^{2/3}(8+h)^{2/3}+ 8^{4/3})\).

Of course, \(\displaystyle (8+ h)^2- 8^2= 64+ 16h+ h^2- 64= 16h+ h^2\) so we can write \(\displaystyle (8+ h)^{2/3}- 8^{2/3}= \frac{h(16+ h)}{((8+h)^{4/3}+ 8^{2/3}(8+h)^{2/3}+ 8^{4/3})}\) and then \(\displaystyle \frac{(8+ h)^{2/3}- 8^{2/3}}{h}= \frac{(16+ h)}{(8+h)^{4/3}+ 8^{2/3}(8+h)^{2/3}+ 8^{4/3}}\) and now it is easy to let h go to 0.