Finding the limit

A

Albi

Junior Member
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May 9, 2020
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Hi everyone, I'm trying to solve this limit but I'm having some issues

[math]\lim_{x \to 0} \left(\frac{\cot(a+2x)-2\cot(a+x)+ \cot(a)}{x^{2}}\right)=[/math][math]=\lim_{x \to 0} \left(\frac{\cot(a+2x)-\cot(a+x)- \cot(a+x) \cot(a)}{x^{2}}\right)[/math]
and after some calculations I've reached this point but I don't think I'm making any progress trying to solve it this way

[math]\lim_{x \to 0} \left(\frac{\cot(a+2x)\cot(a+x)-\cot(a)\cot(a+x)}{\cot(x) \times x^{2}}\right)[/math]
can someone help me continue or suggest another way of solving it
 
Does it look like the second derivative of [imath]\cot[/imath], or is it an optical illusion ?:)
 
Albi said:
Hi everyone, I'm trying to solve this limit but I'm having some issues

[math]\lim_{x \to 0} \left(\frac{\cot(a+2x)-2\cot(a+x)+ \cot(a)}{x^{2}}\right)=[/math][math]=\lim_{x \to 0} \left(\frac{\cot(a+2x)-\cot(a+x)- \cot(a+x) \cot(a)}{x^{2}}\right)[/math]
and after some calculations I've reached this point but I don't think I'm making any progress trying to solve it this way

[math]\lim_{x \to 0} \left(\frac{\cot(a+2x)\cot(a+x)-\cot(a)\cot(a+x)}{\cot(x) \times x^{2}}\right)[/math]
can someone help me continue or suggest another way of solving it
Click to expand...
Have you tried applying L'Hopitals rule?
 
blamocur said:
Does it look like the second derivative of [imath]\cot[/imath], or is it an optical illusion ?:)
Click to expand...
First derivative of 'cot' would be a 'mat' on the floor - and second derivative would be 'bare floor'! Wouldn't it?
 
How does -2cot(a+x) + cot(a) become -cot(a+x) -cot(a+x)cot(a)???

I get that -2cot(a+x) + cot(a) = -cot(a+x) -cot(a+x) + cot(a), but I don't see how the addition becomes multiplication.

Is 8 + 6 = 4 + 4*6?? (I replaced 8=2*4 with 4+4 like you did above)
 
You ask for help. I will give you the best advice/help possible. Don't take calculus until you know algebra and arithmetic. Otherwise the subject will be brutally hard.
 
Steven G said:
You ask for help. I will give you the best advice/help possible. Don't take calculus until you know algebra and arithmetic. Otherwise the subject will be brutally hard.
Click to expand...
I remember trying to explain something about differential equations to students who sincerely believed that [imath]\frac{1}{x}+\frac{1}{y}[/imath] is the same as [imath]\frac{1}{x+y}[/imath]. What a waste of time spending the whole semester on the subject which they had no chance of understanding :(
 
At most community colleges you can take linear algebra after calculus 1. These poor students do not have any chance of learning much in such a class with that limited background. Add in that many are weak in calculus, algebra, trigonometry and arithmetic and you wonder why the instructor even shows up to class.
 
Steven G said:
How does -2cot(a+x) + cot(a) become -cot(a+x) -cot(a+x)cot(a)???

I get that -2cot(a+x) + cot(a) = -cot(a+x) -cot(a+x) + cot(a), but I don't see how the addition becomes multiplication.

Is 8 + 6 = 4 + 4*6?? (I replaced 8=2*4 with 4+4 like you did above)
Click to expand...
It was a typo should be + instead of "*"??, didn't see it, sorry
 
Albi said:
Hi everyone, I'm trying to solve this limit but I'm having some issues

[math]\lim_{x \to 0} \left(\frac{\cot(a+2x)-2\cot(a+x)+ \cot(a)}{x^{2}}\right)=[/math][math]=\lim_{x \to 0} \left(\frac{\cot(a+2x)-\cot(a+x)- \cot(a+x) \cot(a)}{x^{2}}\right)[/math]
and after some calculations I've reached this point but I don't think I'm making any progress trying to solve it this way

[math]\lim_{x \to 0} \left(\frac{\cot(a+2x)\cot(a+x)-\cot(a)\cot(a+x)}{\cot(x) \times x^{2}}\right)[/math]
can someone help me continue or suggest another way of solving it
Click to expand...
If I were to do this problem I would start with:

\(\displaystyle \cot(p + q) = \frac{cot(p) * cot(q) -1}{cot(p) + cot(q)} \)

simplify - then use cot(m) = cos(m)/sin(m) .... and further simplify

Please tell us what you find (in detail) .....
 
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