Question

limit definition of the derivative - homework
determine whether each limit is in the alternate form or the difference quotient, and identify the function, (f(x)) and the value, (c), where the derivative is being evaluated.
alternate form: (f^{prime}(c)=lim_{x
ightarrow c}\frac{f(x)-f(c)}{x - c}) difference quotient: (f^{prime}(c)=lim_{h
ightarrow0}\frac{f(c + h)-f(c)}{h}

1. (lim_{x

ightarrow4}\frac{sqrt{5x}-2sqrt{5}}{x - 4})
(f(x)=)
(c=)

1. (lim_{x

ightarrow7}\frac{ln(7 + h)-ln(7)}{x - 7})
(f(x)=)
(c=)

1. (lim_{h

ightarrow0}\frac{\frac{-2}{(2 + h)^{2}}-(-1)}{h})
(f(x)=)
(c=)

1. (lim_{x

ightarrow - 1}\frac{6x-3x^{3}+3}{x + 1})
(f(x)=)
(c=)

1. (lim_{delta x

ightarrow0}\frac{7-4(-1+delta x)-11}{delta x})
(f(x)=)
(c=)

1. (lim_{x

ightarrow9}\frac{-x^{2}+81}{x - 9})
(f(x)=)
(c=)

1. (lim_{x

ightarrow1}\frac{sec(\frac{pi}{3}x)-2}{x - 1})
(f(x)=)
(c=)

1. (lim_{h

ightarrow0}\frac{2^{3(1 + h)}+5(1 + h)-13}{h})
(f(x)=)
(c=)

Explanation:

Step1: Recall derivative - form identification

Compare given limits with alternate form $f^{\prime}(c)=\lim_{x
ightarrow c}\frac{f(x)-f(c)}{x - c}$ and difference - quotient form $f^{\prime}(c)=\lim_{h
ightarrow0}\frac{f(c + h)-f(c)}{h}$.

Step2: Analyze limit 1

For $\lim_{x
ightarrow4}\frac{\sqrt{5x}-2\sqrt{5}}{x - 4}$, it is in the alternate form. Here $f(x)=\sqrt{5x}$ and $c = 4$.

Step3: Analyze limit 2

The limit $\lim_{x
ightarrow7}\frac{\ln(7 + h)-\ln(7)}{x - 7}$ is incorrect as the variable in the denominator should be $h$ for the difference - quotient or the form is wrong. Assuming it should be $\lim_{h
ightarrow0}\frac{\ln(7 + h)-\ln(7)}{h}$, it is in the difference - quotient form with $f(x)=\ln(x)$ and $c = 7$.

Step4: Analyze limit 3

For $\lim_{h
ightarrow0}\frac{\frac{-2}{(2 + h)^{2}}-(-1)}{h}$, it is in the difference - quotient form. Rewrite it as $\lim_{h
ightarrow0}\frac{f(2 + h)-f(2)}{h}$ where $f(x)=-\frac{2}{x^{2}}$ and $c = 2$.

Step5: Analyze limit 4

For $\lim_{x
ightarrow - 1}\frac{6x-3x^{3}+3}{x + 1}$, it is in the alternate form. We can rewrite it as $\lim_{x
ightarrow - 1}\frac{f(x)-f(-1)}{x+1}$ where $f(x)=6x-3x^{3}+3$ and $c=-1$.

Step6: Analyze limit 5

For $\lim_{\Delta x
ightarrow0}\frac{[7-4(-1+\Delta x)]-11}{\Delta x}$, it is in the difference - quotient form. Rewrite it as $\lim_{\Delta x
ightarrow0}\frac{f(-1+\Delta x)-f(-1)}{\Delta x}$ where $f(x)=7-4x$ and $c=-1$.

Step7: Analyze limit 6

For $\lim_{x
ightarrow9}\frac{-x^{2}+81}{x - 9}$, it is in the alternate form. Rewrite it as $\lim_{x
ightarrow9}\frac{f(x)-f(9)}{x - 9}$ where $f(x)=-x^{2}$ and $c = 9$.

Step8: Analyze limit 7

For $\lim_{x
ightarrow1}\frac{\sec(\frac{\pi}{3}x)-2}{x - 1}$, it is in the alternate form. Rewrite it as $\lim_{x
ightarrow1}\frac{f(x)-f(1)}{x - 1}$ where $f(x)=\sec(\frac{\pi}{3}x)$ and $c = 1$.

Step9: Analyze limit 8

For $\lim_{h
ightarrow0}\frac{2^{3(1 + h)}+5(1 + h)-13}{h}$, it is in the difference - quotient form. Rewrite it as $\lim_{h
ightarrow0}\frac{f(1 + h)-f(1)}{h}$ where $f(x)=2^{3x}+5x$ and $c = 1$.

Answer:

1. Alternate form, $f(x)=\sqrt{5x}$, $c = 4$
2. Assuming correction: Difference - quotient form, $f(x)=\ln(x)$, $c = 7$
3. Difference - quotient form, $f(x)=-\frac{2}{x^{2}}$, $c = 2$
4. Alternate form, $f(x)=6x-3x^{3}+3$, $c=-1$
5. Difference - quotient form, $f(x)=7-4x$, $c=-1$
6. Alternate form, $f(x)=-x^{2}$, $c = 9$
7. Alternate form, $f(x)=\sec(\frac{\pi}{3}x)$, $c = 1$
8. Difference - quotient form, $f(x)=2^{3x}+5x$, $c = 1$