Need help evaluating a limit

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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
 
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Answers and Replies

  • #2
stevendaryl
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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: [itex]\frac{\frac{x^3}{3} + ...}{x^3 + ...}[/itex], where [itex]...[/itex] represents higher-order terms. If you ignore the higher-order terms, what do you get?
 
  • #3
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You're not far from the answer. After cancellations, you have: [itex]\frac{\frac{x^3}{3} + ...}{x^3 + ...}[/itex], where [itex]...[/itex] 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?
 
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  • #4
Samy_A
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You lost a minus sign in the numerator.
 
  • #6
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You lost a minus sign in the numerator.
You are right. It's a typo.
 
  • #7
PeroK
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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##.
 
  • #8
stevendaryl
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You lost a minus sign in the numerator.

No, I think he started with the wrong expansion for tan(x). It should be [itex]tan(x) = x + \frac{x^3}{3} + ...[/itex] not [itex]x - \frac{x^3}{3}[/itex]
 
  • #9
Samy_A
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No, I think he started with the wrong expansion for tan(x). It should be [itex]tan(x) = x + \frac{x^3}{3} + ...[/itex] not [itex]x - \frac{x^3}{3}[/itex]
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##.
 
  • #10
PeroK
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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 :wink:
 
  • #11
Samy_A
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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 :wink:
Sure, the hints he got are strange. This is, as you said, tailor-made(!) for L'Hopital.
 
  • #12
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##\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!}-...}=##
 
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  • #13
PeroK
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##\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}##
 
  • #14
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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.
 
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  • #15
SammyS
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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 .
 

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