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I would like to discuss my solution to a particular limits problem on my exam. I was to evaluate the following limit: [tex]= \lim_{x\to 0}\frac{tan(5x)}{sin(2x)}[/tex]
3. The following is the exact solution I wrote on the exam
My work is in BLACK
My professors corrections are in RED
[tex]= \lim_{x\to 0}\frac{tan(5x)}{sin(2x)}[/tex]
[tex]= \lim_{x\to 0}\frac{\frac{sin(5x)}{cos(5x)}}{sin(2x)}[/tex]
[tex]= \lim_{x\to 0}\frac{sin(5x)}{sin(2x)cos(5x)}[/tex]
[tex]= (\lim_{x\to 0}\frac{sin(5x)}{1}) (\lim_{x\to 0}\frac{1}{sin(2x)}) (\lim_{x\to 0}\frac{1}{cos(5x)})[/tex] --> the middle limit does not exist
(rule:limit of the product is the product of the limit) --> only if the correct limits exist!
[tex]= (\lim_{x\to 0}\frac{5sin(5x)}{5}) (\lim_{x\to 0}\frac{1}{\frac{sin(2x)}{1}}) (\lim_{x\to 0}\frac{1}{\frac{cos(5x)}{1}})[/tex]
[tex]= (5)(1) (\lim_{x\to 0}\frac{1}{\frac{2sin(2x)}{2}}) (\frac{1}{\frac{1}{1}})[/tex]
[tex]= (5)(1) (\frac{1}{2}) (1)[/tex]
[tex]= \frac{5}{2}[/tex] ??
----------------End of work--------------------
So the professor gave me a score of 3/6 on this question, and I believe I deserve more. I don't know where she's going when she wrote that the middle limit does not exist, because this solution is similar to a question she solved herself in class. My solution makes sense to me, and I did get the correct answer. Does anyone else agree with me? I would like to confront the prof and argue this question. Do I have any basis for this argument, or do I really deserve 3/6?
Thanks.
3. The following is the exact solution I wrote on the exam
My work is in BLACK
My professors corrections are in RED
[tex]= \lim_{x\to 0}\frac{tan(5x)}{sin(2x)}[/tex]
[tex]= \lim_{x\to 0}\frac{\frac{sin(5x)}{cos(5x)}}{sin(2x)}[/tex]
[tex]= \lim_{x\to 0}\frac{sin(5x)}{sin(2x)cos(5x)}[/tex]
[tex]= (\lim_{x\to 0}\frac{sin(5x)}{1}) (\lim_{x\to 0}\frac{1}{sin(2x)}) (\lim_{x\to 0}\frac{1}{cos(5x)})[/tex] --> the middle limit does not exist
(rule:limit of the product is the product of the limit) --> only if the correct limits exist!
[tex]= (\lim_{x\to 0}\frac{5sin(5x)}{5}) (\lim_{x\to 0}\frac{1}{\frac{sin(2x)}{1}}) (\lim_{x\to 0}\frac{1}{\frac{cos(5x)}{1}})[/tex]
[tex]= (5)(1) (\lim_{x\to 0}\frac{1}{\frac{2sin(2x)}{2}}) (\frac{1}{\frac{1}{1}})[/tex]
[tex]= (5)(1) (\frac{1}{2}) (1)[/tex]
[tex]= \frac{5}{2}[/tex] ??
----------------End of work--------------------
So the professor gave me a score of 3/6 on this question, and I believe I deserve more. I don't know where she's going when she wrote that the middle limit does not exist, because this solution is similar to a question she solved herself in class. My solution makes sense to me, and I did get the correct answer. Does anyone else agree with me? I would like to confront the prof and argue this question. Do I have any basis for this argument, or do I really deserve 3/6?
Thanks.