identify limits as a derivative

[Sophie]

Question: By identifying each of the following limits as a derivative, find the value of the limit.

\(\displaystyle \L\\{\lim }\limits_{h \to 0} \frac{{\sqrt[3]{{27 + h}} - 3}}{h}\)


\(\displaystyle \L\\{\lim }\limits_{x \to 2} \frac{{x^5 - 32}}{{x - 2}}\)

My problem is I do not really know what is being asked of me, the following is how I think I solve the problem...


\(\displaystyle \L\\\begin{array}{l}
{\lim }\limits_{h \to 0} \frac{{\sqrt[3]{{27 + h}} - 3}}{h} \\
f(h) = \sqrt[3]{{27 + h}} - 3 \\
f(h) = \left( {27 + h} \right)^{1/3} - 3 \\
f'(h) = \frac{1}{3}\left( {27 + h} \right)^{ - 2/3} \\
f'(0) = \frac{1}{3}\left( {27} \right)^{ - 2/3} \\
f'(0) = \frac{1}{{27}} \\
{\lim }\limits_{h \to 0} \frac{{\sqrt[3]{{27 + h}} - 3}}{h} ={\lim }\limits_{h \to 0} \frac{{f(x)}}{x} = {\lim }\limits_{h \to 0} \frac{{f(x) - f(0)}}{{x - 0}} = f'(0) = \frac{1}{{27}} \\
\end{array}\)


\(\displaystyle \L\\\begin{array}{l}
{\lim }\limits_{x \to 2} \frac{{x^5 - 32}}{{x - 2}} \\
f(x) = x^5 - 32 \\
f'(x) = 5x^4 \\
f'(2) = 5(2)^4 \\
f'(2) = 80 \\
{\lim }\limits_{x \to 2} \frac{{x^5 - 32}}{{x - 2}} ={\lim }\limits_{x \to 2} \frac{{f(x)}}{{x - 2}} ={\lim }\limits_{x \to 2} \frac{{f(x) - f(2)}}{{x - 2}} = f'(2) = 80 \\
\end{array}\)


Thanks Sophie

 

[daon]

\(\displaystyle \L\)

 

[mark07]

\(\displaystyle \L f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h} =
\lim_{x\to a} \frac{f(x) - f(a)}{x-a}\)

In (1) \(\displaystyle \L f(x)=\sqrt[3]{x}\) and a=27,

and in (2) \(\displaystyle \L f(x)=x^5\) and a=2.

To find the limits, simply differentiate those functions and evaluate derivatives at a.

 

[Sophie]

Thanks for the help am I right in thinking I was on the right lines, but just not presenting the information correctly.
here is my solution to 1.

\(\displaystyle \L\\\begin{array}{l}
f'(a) ={\lim }\limits_{h \to 0} \frac{{f(a + h) - f(a)}}{h} = {\lim }\limits_{x \to a} \frac{{f(x) - f(a)}}{{x - a}} \\
f(x) = \sqrt[3]{x},a = 27 \\
{\lim }\limits_{x \to 27} \frac{{f(x) - f(27)}}{{x - 27}} \\
\\
f'(x) = \frac{1}{3}(x)^{ - \frac{2}{3}} \\
f'(27) = \frac{1}{3}(27)^{ - \frac{2}{3}} \\
f'(27) = \frac{1}{{27}} \\
\end{array}\)

Here is the solution to 2


\(\displaystyle \L\\\begin{array}{l}
f'(a) = {\lim }\limits_{h \to 0} \frac{{f(a + h) - f(a)}}{h} = {\lim }\limits_{x \to a} \frac{{f(x) - f(a)}}{{x - a}} \\
f(x) = x^5 ,a = 2 \\
{\lim }\limits_{x \to 2} \frac{{f(x) - f(2)}}{{x - 2}} \\
f'(x) = 5x^4 \\
f'(2) = 5(2)^4 \\
f'(2) = 80 \\
\end{array}\)

Thanks Sophie