Finding the limit

[Albi]

Hi everyone, I'm trying to solve this limit but I'm having some issues

$$\lim_{x \to 0} \left(\frac{\cot(a+2x)-2\cot(a+x)+ \cot(a)}{x^{2}}\right)=$$$$=\lim_{x \to 0} \left(\frac{\cot(a+2x)-\cot(a+x)- \cot(a+x) \cot(a)}{x^{2}}\right)$$
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

$$\lim_{x \to 0} \left(\frac{\cot(a+2x)\cot(a+x)-\cot(a)\cot(a+x)}{\cot(x) \times x^{2}}\right)$$
can someone help me continue or suggest another way of solving it

 

[blamocur]

Does it look like the second derivative of $\cot$, or is it an optical illusion ?

 

[Deleted member 4993]

Hi everyone, I'm trying to solve this limit but I'm having some issues

$$\lim_{x \to 0} \left(\frac{\cot(a+2x)-2\cot(a+x)+ \cot(a)}{x^{2}}\right)=$$$$=\lim_{x \to 0} \left(\frac{\cot(a+2x)-\cot(a+x)- \cot(a+x) \cot(a)}{x^{2}}\right)$$
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

$$\lim_{x \to 0} \left(\frac{\cot(a+2x)\cot(a+x)-\cot(a)\cot(a+x)}{\cot(x) \times x^{2}}\right)$$
can someone help me continue or suggest another way of solving it

Have you tried applying L'Hopitals rule?

 

[Deleted member 4993]

Does it look like the second derivative of $\cot$, or is it an optical illusion ?

First derivative of 'cot' would be a 'mat' on the floor - and second derivative would be 'bare floor'! Wouldn't it?

 

[Steven G]

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)

 

[Steven G]

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.

 

[blamocur]

First derivative of 'cot' would be a 'mat' on the floor - and second derivative would be 'bare floor'! Wouldn't it?

Optical illusion after all...

 

[blamocur]

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.

I remember trying to explain something about differential equations to students who sincerely believed that $\frac{1}{x}+\frac{1}{y}$ is the same as $\frac{1}{x+y}$. What a waste of time spending the whole semester on the subject which they had no chance of understanding

 

[Steven G]

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.

 

[Albi]

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)

It was a typo should be + instead of "*"??, didn't see it, sorry

 

[Albi]

Have you tried applying L'Hopitals rule?

I can't use that one, since I'm required to solve it by not using L'Hopital

 

[Deleted member 4993]

Hi everyone, I'm trying to solve this limit but I'm having some issues

$$\lim_{x \to 0} \left(\frac{\cot(a+2x)-2\cot(a+x)+ \cot(a)}{x^{2}}\right)=$$$$=\lim_{x \to 0} \left(\frac{\cot(a+2x)-\cot(a+x)- \cot(a+x) \cot(a)}{x^{2}}\right)$$
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

$$\lim_{x \to 0} \left(\frac{\cot(a+2x)\cot(a+x)-\cot(a)\cot(a+x)}{\cot(x) \times x^{2}}\right)$$
can someone help me continue or suggest another way of solving it

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) .....