Need help evaluating a limit

[lep11]

Homework Statement

Evaluate the limit $\lim_{x\to0} \frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})$ using Taylor's formula. (Hint: $\frac{1}{1+c}=\frac{1-c^2+c^2}{1+c}$ may be useful)

The Attempt at a Solution

I began by substituting $tanx$ with $x+\frac{x^3}{3}+x^3ε(x)$, where ε tends to zero as x approaches 0.

$\frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})=\frac{1}{x}(\frac{x-tanx}{xtanx})=\frac{1}{x}(\frac{x-x-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=\frac{1}{x}(\frac{-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^2(x+\frac{x^3}{3}+x^3ε(x))}=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^3+\frac{x^4}{3}+x^4ε(x))}=\frac{-\frac{1}{3}-ε(x))}{1+\frac{x}{3}+xε(x))}$ →-1/3 as x→0

 

[stevendaryl]

Homework Statement

Evaluate the limit $\lim_{x\to0} \frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})$ using Taylor's formula. (Hint: $\frac{1}{1+c}=\frac{1-c^2+c^2}{1+c}$ may be useful)

The Attempt at a Solution

substituting $tanx$ with $x-\frac{x^3}{3}+x^3ε(x)$, where ε tends to zero as x approaches 0.

$\frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})=\frac{1}{x}(\frac{x-tanx}{xtanx})=\frac{1}{x}(\frac{x-x-\frac{x^3}{3}+x^3ε(x))}{x(x-\frac{x^3}{3}+x^3ε(x))})=\frac{1}{x}(\frac{\frac{x^3}{3}+x^3ε(x))}{x(x-\frac{x^3}{3}+x^3ε(x))})$

You're not far from the answer. After cancellations, you have: $\frac{\frac{x^3}{3} + ...}{x^3 + ...}$, where $...$ represents higher-order terms. If you ignore the higher-order terms, what do you get?

 

[lep11]

You're not far from the answer. After cancellations, you have: $\frac{\frac{x^3}{3} + ...}{x^3 + ...}$, where $...$ represents higher-order terms. If you ignore the higher-order terms, what do you get?

It's 1/3. But where to use the hint I am given?

 

[Samy_A]

You lost a minus sign in the numerator.

 

[stevendaryl]

It's 1/3. But where to use the hint I am given?

I don't get that, either.

 

[lep11]

You lost a minus sign in the numerator.

You are right. It's a typo.

 

[PeroK]

Homework Statement

Evaluate the limit $\lim_{x\to0} \frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})$ using Taylor's formula. (Hint: $\frac{1}{1+c}=\frac{1-c^2+c^2}{1+c}$ may be useful)

The Attempt at a Solution

I began by substituting $tanx$ with $x-\frac{x^3}{3}+x^3ε(x)$, where ε tends to zero as x approaches 0.

$\frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})=\frac{1}{x}(\frac{x-tanx}{xtanx})=\frac{1}{x}(\frac{x-x-\frac{x^3}{3}+x^3ε(x))}{x(x-\frac{x^3}{3}+x^3ε(x))})=\frac{1}{x}(\frac{\frac{x^3}{3}+x^3ε(x))}{x(x-\frac{x^3}{3}+x^3ε(x))})=\frac{\frac{x^3}{3}+x^3ε(x))}{x^2(x-\frac{x^3}{3}+x^3ε(x))}=\frac{\frac{x^3}{3}+x^3ε(x))}{x^3-\frac{x^4}{3}+x^4ε(x))}$??

I think you've got the wrong series for $tan(x)$. Check your coefficients. Also, if you are going to use the Taylor series, you should use the series for $1/tan(x)$ by applying the binomial expansion to the series for $tan(x)$ or using the series for $cot(x)$.

That said, this one looks tailor-made(!) for L'Hopital, using $tan = sin/cos$.

 

[stevendaryl]

You lost a minus sign in the numerator.

No, I think he started with the wrong expansion for tan(x). It should be $tan(x) = x + \frac{x^3}{3} + ...$ not $x - \frac{x^3}{3}$

 

[Samy_A]

No, I think he started with the wrong expansion for tan(x). It should be $tan(x) = x + \frac{x^3}{3} + ...$ not $x - \frac{x^3}{3}$

Yes, he has a wrong Taylor series. In the denominator, that should have been $+\frac{x^3}{3}$. But in the numerator, he expands $x-\tan x$, so the $\frac{x^3}{3}$ gets a minus sign.
When taking the limit, the $x³$ term resulting from the $\tan$ series in the denominator is not important, but the one in the numerator is.
The correct limit is $-\frac13$.

 

[PeroK]

My advice would be to do it the easy way using L'Hopital so you know what the answer is, then do it the hard way using Taylor series

 

[Samy_A]

My advice would be to do it the easy way using L'Hopital so you know what the answer is, then do it the hard way using Taylor series

Sure, the hints he got are strange. This is, as you said, tailor-made(!) for L'Hopital.

 

[lep11]

$\frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})=\frac{1}{x}(\frac{x-tanx}{xtanx})=\frac{1}{x}(\frac{x-x-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=...=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^3+\frac{x^4}{3}+x^4ε(x))}=\frac{-\frac{1}{3}-ε(x))}{1+\frac{x}{3}+xε(x))}$ →-1/3 as x→0

Looks like I didn't need the hint. However, I think I'm supposed to derive the Taylor formula I used for the problem. We have covered expansions for sinx and cos x in class. So $tanx=\frac{sinx}{cosx}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{1-\frac{x^2}{2!}+\frac{x^4}{4!}-...}=$

 

[PeroK]

$\frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})=\frac{1}{x}(\frac{x-tanx}{xtanx})=\frac{1}{x}(\frac{x-x-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=...=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^3+\frac{x^4}{3}+x^4ε(x))}=\frac{-\frac{1}{3}-ε(x))}{1+\frac{x}{3}+xε(x))}$ →-1/3 as x→0

Looks like I didn't need the hint. However, I think I'm supposed to derive the Taylor formula I used for the problem. We have covered expansions for sinx and cos x in class. So $tanx=\frac{sinx}{cosx}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{1-\frac{x^2}{2!}+\frac{x^4}{4!}-...}=$

I would go for: $1/tanx = cosx/sinx$

You might as well make things easy for yourself!

Use the binomial for $1/sinx = (x-\frac{x^3}{3!}+\frac{x^5}{5!}- \dots)^{-1} = (1/x)(1-\frac{x^2}{3!}+\frac{x^4}{5!}- \dots)^{-1}$

 

[lep11]

I actually derived the expansion for $\tan x$ from the definition of taylor formula. $f'(x)=D\tan x=1+\tan^2 x$, $f'(0)=1$
$f''(x)=2\tan x(1+\tan^2 x)$, $f''(0)=0$ etc. and got $\tan x=x+\frac{x^3}{3}+x^3ε(x)$. I think this is maybe shorter and nicer than deriving it by long division.

 

[SammyS]

I would go for: $1/ \tan x = \cos x / \sin x$

I thought I had posted something on this a day or two ago. Using the first of Perok's suggestions, then common denominator, etc. :

$\displaystyle \ \frac{1}{x}\left(\frac{1}{\tan x}-\frac{1}{x}\right) \$

$\displaystyle \ =\frac{x \cos x - \sin x}{x^2\sin x} \$

Now use the Taylor expansions for sin x and cos x .