limit definition of the derivative - homework\ndetermine whether each limit is in the alternate form or the…

limit definition of the derivative - homework\ndetermine 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.\nalternate form: (f^{prime}(c)=lim_{x\rightarrow c}\frac{f(x)-f(c)}{x - c}) difference quotient: (f^{prime}(c)=lim_{h\rightarrow0}\frac{f(c + h)-f(c)}{h}\n1. (lim_{x\rightarrow4}\frac{sqrt{5x}-2sqrt{5}}{x - 4})\n(f(x)=)\n(c=)\n2. (lim_{x\rightarrow7}\frac{ln(7 + h)-ln(7)}{x - 7})\n(f(x)=)\n(c=)\n3. (lim_{h\rightarrow0}\frac{\frac{-2}{(2 + h)^{2}}-(-1)}{h})\n(f(x)=)\n(c=)\n4. (lim_{x\rightarrow - 1}\frac{6x-3x^{3}+3}{x + 1})\n(f(x)=)\n(c=)\n5. (lim_{delta x\rightarrow0}\frac{7-4(-1+delta x)-11}{delta x})\n(f(x)=)\n(c=)\n6. (lim_{x\rightarrow9}\frac{-x^{2}+81}{x - 9})\n(f(x)=)\n(c=)\n7. (lim_{x\rightarrow1}\frac{sec(\frac{pi}{3}x)-2}{x - 1})\n(f(x)=)\n(c=)\n8. (lim_{h\rightarrow0}\frac{2^{3(1 + h)}+5(1 + h)-13}{h})\n(f(x)=)\n(c=)
Answer
Explanation:
Step1: Recall derivative - form identification
Compare given limits with alternate form $f^{\prime}(c)=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x - c}$ and difference - quotient form $f^{\prime}(c)=\lim_{h\rightarrow0}\frac{f(c + h)-f(c)}{h}$.
Step2: Analyze limit 1
For $\lim_{x\rightarrow4}\frac{\sqrt{5x}-2\sqrt{5}}{x - 4}$, it is in the alternate form. Here $f(x)=\sqrt{5x}$ and $c = 4$.
Step3: Analyze limit 2
The limit $\lim_{x\rightarrow7}\frac{\ln(7 + h)-\ln(7)}{x - 7}$ is incorrect as the variable in the denominator should be $h$ for the difference - quotient or the form is wrong. Assuming it should be $\lim_{h\rightarrow0}\frac{\ln(7 + h)-\ln(7)}{h}$, it is in the difference - quotient form with $f(x)=\ln(x)$ and $c = 7$.
Step4: Analyze limit 3
For $\lim_{h\rightarrow0}\frac{\frac{-2}{(2 + h)^{2}}-(-1)}{h}$, it is in the difference - quotient form. Rewrite it as $\lim_{h\rightarrow0}\frac{f(2 + h)-f(2)}{h}$ where $f(x)=-\frac{2}{x^{2}}$ and $c = 2$.
Step5: Analyze limit 4
For $\lim_{x\rightarrow - 1}\frac{6x-3x^{3}+3}{x + 1}$, it is in the alternate form. We can rewrite it as $\lim_{x\rightarrow - 1}\frac{f(x)-f(-1)}{x+1}$ where $f(x)=6x-3x^{3}+3$ and $c=-1$.
Step6: Analyze limit 5
For $\lim_{\Delta x\rightarrow0}\frac{[7-4(-1+\Delta x)]-11}{\Delta x}$, it is in the difference - quotient form. Rewrite it as $\lim_{\Delta x\rightarrow0}\frac{f(-1+\Delta x)-f(-1)}{\Delta x}$ where $f(x)=7-4x$ and $c=-1$.
Step7: Analyze limit 6
For $\lim_{x\rightarrow9}\frac{-x^{2}+81}{x - 9}$, it is in the alternate form. Rewrite it as $\lim_{x\rightarrow9}\frac{f(x)-f(9)}{x - 9}$ where $f(x)=-x^{2}$ and $c = 9$.
Step8: Analyze limit 7
For $\lim_{x\rightarrow1}\frac{\sec(\frac{\pi}{3}x)-2}{x - 1}$, it is in the alternate form. Rewrite it as $\lim_{x\rightarrow1}\frac{f(x)-f(1)}{x - 1}$ where $f(x)=\sec(\frac{\pi}{3}x)$ and $c = 1$.
Step9: Analyze limit 8
For $\lim_{h\rightarrow0}\frac{2^{3(1 + h)}+5(1 + h)-13}{h}$, it is in the difference - quotient form. Rewrite it as $\lim_{h\rightarrow0}\frac{f(1 + h)-f(1)}{h}$ where $f(x)=2^{3x}+5x$ and $c = 1$.
Answer:
- Alternate form, $f(x)=\sqrt{5x}$, $c = 4$
- Assuming correction: Difference - quotient form, $f(x)=\ln(x)$, $c = 7$
- Difference - quotient form, $f(x)=-\frac{2}{x^{2}}$, $c = 2$
- Alternate form, $f(x)=6x-3x^{3}+3$, $c=-1$
- Difference - quotient form, $f(x)=7-4x$, $c=-1$
- Alternate form, $f(x)=-x^{2}$, $c = 9$
- Alternate form, $f(x)=\sec(\frac{\pi}{3}x)$, $c = 1$
- Difference - quotient form, $f(x)=2^{3x}+5x$, $c = 1$