Limits problem

[Precursor]

I would like to discuss my solution to a particular limits problem on my exam. I was to evaluate the following limit: = \lim_{x\to 0}\frac{tan(5x)}{sin(2x)}

3. The following is the exact solution I wrote on the exam
My work is in BLACK
My professors corrections are in RED

= \lim_{x\to 0}\frac{tan(5x)}{sin(2x)}

= \lim_{x\to 0}\frac{\frac{sin(5x)}{cos(5x)}}{sin(2x)}

= \lim_{x\to 0}\frac{sin(5x)}{sin(2x)cos(5x)}

= (\lim_{x\to 0}\frac{sin(5x)}{1}) (\lim_{x\to 0}\frac{1}{sin(2x)}) (\lim_{x\to 0}\frac{1}{cos(5x)}) --> the middle limit does not exist

(rule:limit of the product is the product of the limit) --> only if the correct limits exist!!

= (\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}})

= (5)(1) (\lim_{x\to 0}\frac{1}{\frac{2sin(2x)}{2}}) (\frac{1}{\frac{1}{1}})

= (5)(1) (\frac{1}{2}) (1)

= \frac{5}{2} ??

----------------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.

[FirstYearGrad]

Certainly you got the correct answer, but the professor is correct that the middle limit does not exist (check from both sides as approaching 0). I don't think it will be productive to argue subjective points with the professor.

[Precursor]

Then what must I have instead of the middle limit?

Besides, I don't see why 3 mark are deducted. She was very fair in other questions, so it just seems really odd that 3 marks are deducted for making an error in one step.

[FirstYearGrad]

I think you were given half credit because you failed to see the whole "trick" while not introducing any "evil" like the professor pointed out. I don't know exactly what you know at this point in the course, but it possibly involves multiplying the limit by \frac{10x}{10x} to yield:

= (\lim_{x\to 0}\frac{5sin(5x)}{5x}) (\lim_{x\to 0}\frac{2x}{2sin(2x)}) (\lim_{x\to 0}\frac{1}{cos(5x)})

edit: whoops minor mistake.

[Precursor]

So how is your middle limit different than mine? Does it not also exist?

You see, all I did was separate = \lim_{x\to 0}\frac{sin(5x)}{sin(2x)cos(5x)} into 3 limits.

[FirstYearGrad]

Mine does exist. The reason why yours doesn't is that if you observe the value from infinity to 0 the limit tends toward infinity, whereas if you observe the value from -infinity to 0 the limit tends toward -infinity. Mine, however, does not, as the addition of x to the numerator causes the signs between the numerator and denominator to always cancel and therefore the limit exists as it converges to the same value from either side.

[Precursor]

So writing \lim_{x\to 0}\frac{1}{sin(2x)} alone is incorrect?

[FirstYearGrad]

Right.

[Precursor]

Yes I see how the prof is correct in this now, but I still don't believe my errors account for half of the solution.

Does anybody else have something to add?

[FirstYearGrad]

Well you can take it up with the professor, but definitely keep in mind that a more craftily-made problem could have caused your thinking to give the incorrect solution. For example, the limit of \lim_{x \to 0}\frac{sin(x)}{sin(2x)sin(3x)}

[DR13]

= (\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}})

= (5)(1) (\lim_{x\to 0}\frac{1}{\frac{2sin(2x)}{2}}) (\frac{1}{\frac{1}{1}})

= (5)(1) (\frac{1}{2}) (1)

= \frac{5}{2} ??



Also:
when you take lim(x-0) of 5sin(5x)/5 you get 0. not 5 as you stated. this is due to the fact that sin(ax)=0. Furthermore lim(x-0) of 1/((2sin(2x))/2) is undefined because the fraction would end up being undefined because it would be equal to 1/(0/1) when the limit is applied (again because sin(ax)=0). Then you miraculously arrive at the correct answer.

[Precursor]

So I seem to have 2 errors in my solution. But shouldn't I get at least one more mark for getting the right answer?

[DR13]

I think the problem is that your answer comes from nowhere. You have incorrect math that somehow arrives at the correct answer. In my opinion, you should be happy with your 3/6...