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Thread: Continuity in Composite functions

  1. #1
    Senior Member pankaj's Avatar
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    Continuity in Composite functions

    If $\displaystyle \lim_{x\to a}g(x)$ exists then is it proper to say that
    $\displaystyle
    \lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))
    $provided f(x) is continuous at x=a.

    I read this somewhere and do not believe it to be authentic.
    I have come across many examples which appear to support the above statement but then there are other examples which do not support this.

    If it is true then is there a rigorous proof in support of this.

    One example that I have is as follows:
    $\displaystyle f(x)=[x]$
    $\displaystyle
    g(x)=\frac{sinx}{x}
    $
    In this case is
    $\displaystyle
    \lim_{x\to 0}f(g(x))=f(\lim_{x\to 0}g(x))
    $

    Another example that i have is:
    $\displaystyle f(x)=[x]$
    $\displaystyle
    g(x)=\frac{tanx}{x}
    $
    In this case is
    $\displaystyle
    \lim_{x\to 0}f(g(x))=f(\lim_{x\to 0}g(x))
    $
    Would be really grateful if someone verifies this for me
    Thanx in advance
    Also i must mention that [x] denotes the greatest integer less than or equal to x
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  2. #2
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    Quote Originally Posted by pankaj View Post
    If $\displaystyle \lim_{x\to a}g(x)$ exists then is it proper to say that
    $\displaystyle
    \lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))
    $provided f(x) is continuous at x=a.
    You got the theorem wrong.
    If $\displaystyle \lim_{x\to a}g(x)=L$ and $\displaystyle f$ is continoust at $\displaystyle L$ then $\displaystyle \lim_{x\to a}f(g(x)) = f(L)$.

    We will prove it. Let $\displaystyle \epsilon > 0$.
    Since $\displaystyle f$ is continous at $\displaystyle L$ it means there is $\displaystyle \delta > 0$ such that $\displaystyle |f(y) - f(L)|<\epsilon$ for $\displaystyle |y-L| < \delta$. Now $\displaystyle \lim_{x\to a}g(x) = L$ thus there is $\displaystyle \eta > 0$ such that if $\displaystyle 0<|x-a|<\eta$ implies $\displaystyle |g(x) - L| < \delta$. Putting these together we see that if $\displaystyle 0<|x-a|<\eta$ then $\displaystyle |f(g(x))-f(L)| < \epsilon$. Thus, $\displaystyle \lim_{x\to a}f(g(x)) = f(L)$.
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  3. #3
    Senior Member pankaj's Avatar
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    O.K.
    Now we know that $\displaystyle 0<\frac{\sin x}{x}<1$ when x is in the neighbourhood of 0.
    Therefore, $\displaystyle [\frac{\sin x}{x}]=0$ which means
    $\displaystyle
    \lim_{x\to 0}[\frac{\sin x}{x}]=0
    $

    Now,since
    $\displaystyle
    \lim_{x\to 0}\frac{\sin x}{x}=1
    $

    Should
    $\displaystyle
    [\lim_{x\to 0}\frac{\sin x}{x}]=1
    $
    or should
    $\displaystyle
    [\lim_{x\to 0}\frac{\sin x}{x}]=0
    $
    This is what I want to verify.

    Also it is known that $\displaystyle 1<\frac{\tan x}{x}<2$ in neighbourhood of 0 and
    $\displaystyle
    \lim_{x\to 0}\frac{\tan x}{x}=1
    $

    Therefore
    $\displaystyle
    \lim_{x\to 0}[\frac{\tan x}{x}]=1
    $
    What should be value of
    $\displaystyle
    [\lim_{x\to 0}\frac{\tan x}{x}]
    $
    Is it same as
    $\displaystyle
    \lim_{x\to 0}[\frac{\tan x}{x}]
    $

    This is what I want to verify
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  4. #4
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    Look at the graphs of the two functions.
    Then think in therms of the floor function.
    It should be clear that $\displaystyle \lim _{x \to 0} \left( {\left\lfloor {\frac{{\sin (x)}}{x}} \right\rfloor } \right) = 0\,\& \,\lim _{x \to 0} \left( {\left\lfloor {\frac{{\tan (x)}}{x}} \right\rfloor } \right) = 1$.
    Attached Thumbnails Attached Thumbnails Continuity in Composite functions-floor.gif  
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  5. #5
    Senior Member pankaj's Avatar
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    Agreed.
    What I want to know is that in these two examples
    Is $\displaystyle \lim_{x\to 0}[\frac{\sin x}{x}] =[\lim_{x\to 0}\frac{\sin x}{x}]$
    and
    is $\displaystyle \lim_{x\to 0}[\frac{\tan x}{x}] =[\lim_{x\to 0}\frac{\tan x}{x}]$

    Would appreciate if one describes in detail
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  6. #6
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    Well, have you looked at the graphs.
    Attached Thumbnails Attached Thumbnails Continuity in Composite functions-floor1.gif  
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  7. #7
    Senior Member pankaj's Avatar
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    I get your point
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  8. #8
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    you don't need to sketch a graph. choose $\displaystyle 0 < x < \frac{\pi}{2}.$ then if you apply mean value theorem for $\displaystyle f(t)=\sin t$ on the interval $\displaystyle [0,x],$ then you get $\displaystyle 0 < \frac{\sin x}{x} < 1.$

    thus, since $\displaystyle \frac{\sin x}{x}$ is an even function, for any non-zero $\displaystyle x$ in the interval $\displaystyle [-\pi/2, \pi/2]$ we must have: $\displaystyle 0< \frac{\sin x}{x} < 1,$ which gives us: $\displaystyle \lfloor{\frac{\sin x}{x} \rfloor}=0,$ for any $\displaystyle x \neq 0$ in

    that interval. thus $\displaystyle \lim_{x\to0} \lfloor{\frac{\sin x}{x} \rfloor}= \lim_{x\to 0} 0 = 0.$ but $\displaystyle \lfloor{\lim_{x\to0} \frac{\sin x}{x} \rfloor}=\lfloor{1 \rfloor}=1.$ a similar argument works for $\displaystyle \frac{\tan x}{x}.$
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  9. #9
    Senior Member pankaj's Avatar
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    This is precisely what I wanted to know that while evaluating
    $\displaystyle
    [\lim_{x\to 0}\frac{\sin x}{x}]
    $
    we are required to first find out
    $\displaystyle
    \lim_{x\to 0}\frac{\sin x}{x}
    $
    and then apply rule of greatest integer function(I would want to know the reason.Is there a rigorous proof?)
    Or should we not say that since $\displaystyle 0<\frac{\sin x}{x}<1$ the greatest integer should anyway be 0 whether we find
    $\displaystyle
    [\lim_{x\to 0}\frac{\sin x}{x}]
    $
    or we find
    $\displaystyle
    \lim_{x\to 0}[\frac{\sin x}{x}]
    $

    I am perplexed because in case of $\displaystyle \frac{\tan x}{x}$ the answer for
    $\displaystyle
    [\lim_{x\to 0}\frac{\tan x}{x}]=1
    $

    $\displaystyle
    \lim_{x\to 0}[\frac{\tan x}{x}]=1
    $
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  10. #10
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    Quote Originally Posted by pankaj View Post
    I am perplexed because in case of $\displaystyle \frac{\tan x}{x}$ the answer for
    $\displaystyle
    [\lim_{x\to 0}\frac{\tan x}{x}]=1
    $

    $\displaystyle
    \lim_{x\to 0}[\frac{\tan x}{x}]=1
    $
    there's this general fact: suppose $\displaystyle \lim_{x\to a}f(x)=b,$ and $\displaystyle g$ is right continuous at $\displaystyle x=b.$ if $\displaystyle f(x) \geq b,$ in some neighbourhood of $\displaystyle x=a,$ then $\displaystyle \lim_{x\to a}g(f(x))=g(\lim_{x\to a}f(x))=g(b).$

    the same result holds if we replace right with left and $\displaystyle f(x) \geq b$ with $\displaystyle f(x) \leq b.$ i'll prove the claim for the right continuous case: let $\displaystyle \epsilon > 0$. then since $\displaystyle g$ is right continous at $\displaystyle b,$

    there exists $\displaystyle \delta_1 > 0$ such that $\displaystyle |g(t)-g(b)| < \epsilon$ whenever $\displaystyle 0 \leq t - b < \delta_1.$ also, since $\displaystyle \lim_{x\to a}f(x)=b,$ there exists $\displaystyle \delta_2 > 0$ such that $\displaystyle |f(x)-b| < \delta_1$ whenever $\displaystyle 0<|x-a| < \delta_2.$

    finally we know that there exists $\displaystyle \delta_3 > 0$ such that $\displaystyle f(x) \geq b$ whenever $\displaystyle 0 < |x-a| < \delta_3.$ now let $\displaystyle \delta=\min(\delta_2,\delta_3)$ and $\displaystyle 0 < |x-a| < \delta.$ then: $\displaystyle 0 \leq f(x)-b=|f(x)-b| < \delta_1.$

    thus we must have: $\displaystyle |g(f(x))-g(b)| < \epsilon,$ which proves that $\displaystyle \lim_{x\to a}g(f(x))=g(b). \ \boxed{\text{NCA}}$


    as an example you can check your question: the function $\displaystyle g(x)=\lfloor{x \rfloor}$ is right continuous at x = 0 and $\displaystyle f(x)=\frac{\tan x}{x} \geq 1$ in some neighbourhood of x = 0. that's why you got that

    result. if instead you take $\displaystyle g$ to be the ceiling function, which is left continuous at x = 0, then you'll get the same result but for the function $\displaystyle f(x)=\frac{\sin x}{x}.$
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