Results 1 to 4 of 4

Thread: Trigonometric limits - question involving tan.

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    83

    Trigonometric limits - question involving tan.

    I am having trouble answering this question:

    "Using $\displaystyle tanx = \frac{sinx}{cosx} $ evaluate *limit as x approaches 0* $\displaystyle \frac{tanx - sinx}{xcosx} $"

    My main problem, I think, is getting rid of the x on the bottom, I do not know how to go about doing it.

    My solution so far:
    note:*the "limit as x approaches 0" I don't know how to input on latex, so pretend it is there before all these.*

    $\displaystyle = \frac{\frac{sinx}{cosx} - sinx}{xcosx} $
    $\displaystyle = \frac{\frac{sinx - sinxcosx}{cosx}}{xcosx} $
    $\displaystyle = \frac {sinx - sinx}{xcosx} $

    Don't know what to do from here. I know the answer is zero, but I do not know how to get there from what I have. Help please.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    May 2009
    From
    Zagreb
    Posts
    65
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, Kakariki!

    We're expected to know: .$\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$


    I am having trouble answering this question:

    Using $\displaystyle \tan x \:=\: \frac{\sin x}{\cos x}$, evaluate: .$\displaystyle \lim_{x\to0}\frac{\tan x - \sin x}{x\cos x} $

    My main problem, I think, is getting rid of the x on the bottom, I do not know how to go about doing it.

    My solution so far:

    $\displaystyle \lim_{x\to0} \frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x} $

    . . $\displaystyle =\;\lim_{x\to0}\frac{\frac{\sin x - \sin x\cos x}{\cos x}}{x\cos x} $

    . . $\displaystyle = \;\lim_{x\to0}\frac {\sin x - \sin x}{x\cos x} $ . . . . wrong!

    We have: .$\displaystyle \frac{\tan x - \sin x}{x\cos x} \;=\;\frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x}$

    Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\quad \frac{\cos x\left(\frac{\sin x}{\cos x} - \sin x\right)}{\cos x(x\cos x)} \;=\;\frac{\sin x - \sin x\cos x}{x\cos^2\!x} \;=\;\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}$


    $\displaystyle \text{Take the limit: }\;\lim_{x\to0}\bigg[\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}\bigg] \;=\;\underbrace{\lim_{x\to0}\left(\frac{\sin x}{x}\right)}_1\cdot\underbrace{\lim_{x\to0}\left( \frac{1-\cos x}{\cos^2\!x}\right)}_{\frac{1-1}{1^2}}$

    . . . . . . . . . $\displaystyle = \;1\cdot0 \;=\;0 $

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2009
    Posts
    83
    Quote Originally Posted by Soroban View Post
    Hello, Kakariki!

    We're expected to know: .$\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$



    We have: .$\displaystyle \frac{\tan x - \sin x}{x\cos x} \;=\;\frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x}$

    Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\quad \frac{\cos x\left(\frac{\sin x}{\cos x} - \sin x\right)}{\cos x(x\cos x)} \;=\;\frac{\sin x - \sin x\cos x}{x\cos^2\!x} \;=\;\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}$


    $\displaystyle \text{Take the limit: }\;\lim_{x\to0}\bigg[\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}\bigg] \;=\;\underbrace{\lim_{x\to0}\left(\frac{\sin x}{x}\right)}_1\cdot\underbrace{\lim_{x\to0}\left( \frac{1-\cos x}{\cos^2\!x}\right)}_{\frac{1-1}{1^2}}$

    . . . . . . . . . $\displaystyle = \;1\cdot0 \;=\;0 $

    Okay, that makes sense. How did you know to multiply by $\displaystyle \frac{cosx}{cosx} $ though? Is it because you can't have a 0 in the denominator, but you can in the numerator?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving trigonometric equations involving arctan?
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: Oct 3rd 2011, 04:29 AM
  2. Replies: 3
    Last Post: Jun 17th 2011, 10:17 AM
  3. Question about Limits involving infinity
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 28th 2010, 07:48 AM
  4. Limits Involving Trigonometric Functions
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Feb 1st 2010, 07:59 PM
  5. Replies: 3
    Last Post: Oct 5th 2007, 04:22 AM

Search Tags


/mathhelpforum @mathhelpforum