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Thread: discontinuity

  1. #1
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    discontinuity

    f(x)=2sinx-cosx
    -----------
    1+tanx

    In how many points in the interval [0,2pi] is the function above discontinious ?

    I say 3 points book says 2.

    pi/2:tangent is undefined
    tan135:denominator becomes 0
    tan315: denominator becomes 0.
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    Re: discontinuity

    The function is continuous at $x= \frac{\pi}{2}$. Look at the limits as $x \rightarrow \frac{\pi}{2}$ from the left and from the right.
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    Re: discontinuity

    -infinite from the right and +infinite from the left
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    Junior Member Ahri's Avatar
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    Re: discontinuity

    Then what can you say about limit of $f(x)$ as $x \rightarrow \frac{\pi}{2}$
    Last edited by Ahri; May 30th 2017 at 10:58 AM.
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    Re: discontinuity

    It has no limit.
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    Re: discontinuity

    It tends to zero from both sides...

    From the left ($\pi/2$), $\tan x $ tends to $+ \infty$. From the right, $\tan x $ tends to $-\infty$.

    So $$\lim_{x \rightarrow \frac{\pi}{2}^-} f(x)= \frac{2}{1+\infty}=0$$
    and
    $$\lim_{x \rightarrow \frac{\pi}{2}^+} f(x)= \frac{2}{1-\infty}=0$$

    So from this we can say that $$\lim_{x \rightarrow \frac{\pi}{2}} f(x) =0$$.

    Now a function is continuous at a point $a$ if $$\lim_{x \rightarrow a} f(x) = f(a)$$

    Can you re-write $f(x)$ so that you can evaluate it at $x=\pi /2$? Hint: Use $\tan x = \frac{ \sin x}{\cos x}$ and write $1+ \tan x = \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}$ .
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    Re: discontinuity

    Kastamonu, Ahri is offering a different function that is continuous at $\frac{\pi}{2}$. The function you are given is undefined at that point, so you are correct that is it discontinuous. The definition of continuous requires the function be defined AND that the limit from the left and from the right be equal to the value defined.

    There will be four points of discontinuity on the given interval.

    $\dfrac{\pi}{2},\dfrac{3\pi}{4},\dfrac{3\pi}{2}, \dfrac{7\pi}{4}$
    Last edited by SlipEternal; May 30th 2017 at 12:37 PM.
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    Junior Member Ahri's Avatar
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    Re: discontinuity

    Quote Originally Posted by SlipEternal View Post
    Kastamonu, Ahri is offering a different function that is continuous at $\frac{\pi}{2}$. The function you are given is undefined at that point, so you are correct that is it discontinuous. The definition of continuous requires the function be defined AND that the limit from the left and from the right be equal to the value defined.

    There will be three points of discontinuity on the given interval.
    I was going on $$f\left(x\right)=\frac{2\sin x - \cos x}{1+ \tan x}=\frac{2\sin x - \cos x}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}=\frac{2\sin x - \cos x}{\frac{1}{\cos x}\left(\cos x + \sin x\right)}=\frac{\cos x \left(2\sin x - \cos x\right)}{\cos x + \sin x} $$

    which can be evaluated at $x= \frac{\pi}{2}$
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    Re: discontinuity

    Quote Originally Posted by Ahri View Post
    I was going on $$f\left(x\right)=\frac{2\sin x - \cos x}{1+ \tan x}=\frac{2\sin x - \cos x}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}=\frac{2\sin x - \cos x}{\frac{1}{\cos x}\left(\cos x + \sin x\right)}=\frac{\cos x \left(2\sin x - \cos x\right)}{\cos x + \sin x} $$

    which can be evaluated at $x= \frac{\pi}{2}$
    That would be like saying the rational function $\dfrac{x^2}{x}$ can be evaluated at 0.
    Thanks from Ahri
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    Junior Member Ahri's Avatar
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    Re: discontinuity

    It's been a while since I've studied these things, apologies then .
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    Re: discontinuity

    f(x)=(2sinx-cosx)/(1+tanx)
    The function is like that.
    tan135:denominator becomes 0
    tan315: denominator becomes 0.
    At these points it is discontinious
    But I am not sure about 90 and 270. Tangent is undefined at these points.
    According to the book I am wrong but Ahri is right.
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    Re: discontinuity

    Quote Originally Posted by kastamonu View Post
    f(x)=(2sinx-cosx)/(1+tanx)
    The function is like that.
    tan135:denominator becomes 0
    tan315: denominator becomes 0.
    At these points it is discontinious
    But I am not sure about 90 and 270. Tangent is undefined at these points.
    According to the book I am wrong but Ahri is right.
    What is the definition of continuity according to your book?
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    Re: discontinuity

    f:A--->R
    a is element of A
    There is a limit when x goes to a and this limit is equal to f(a).

    If we put our finger on one end of the function and go to the other end without removing our finger then we can say that the function is continious.
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  14. #14
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    Re: discontinuity

    Perfect, since the function does not exist, it cannot be continuous when tangent does not exist.
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  15. #15
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    Re: discontinuity

    Sorry if I confused you Kastamonu, I'll be more careful next time .
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