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Math Help - Continuity in Composite functions

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

    If \lim_{x\to a}g(x) exists then is it proper to say that
     <br />
\lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))<br />
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:
    f(x)=[x]
     <br />
g(x)=\frac{sinx}{x}<br />
    In this case is
     <br />
\lim_{x\to 0}f(g(x))=f(\lim_{x\to 0}g(x))<br />

    Another example that i have is:
    f(x)=[x]
     <br />
g(x)=\frac{tanx}{x}<br />
    In this case is
     <br />
\lim_{x\to 0}f(g(x))=f(\lim_{x\to 0}g(x))<br />
    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 \lim_{x\to a}g(x) exists then is it proper to say that
     <br />
\lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))<br />
provided f(x) is continuous at x=a.
    You got the theorem wrong.
    If \lim_{x\to a}g(x)=L and f is continoust at L then \lim_{x\to a}f(g(x)) = f(L).

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

    Now,since
     <br />
\lim_{x\to 0}\frac{\sin x}{x}=1<br />

    Should
     <br />
[\lim_{x\to 0}\frac{\sin x}{x}]=1 <br />
    or should
     <br />
[\lim_{x\to 0}\frac{\sin x}{x}]=0 <br />
    This is what I want to verify.

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

    Therefore
     <br />
\lim_{x\to 0}[\frac{\tan x}{x}]=1 <br />
    What should be value of
     <br />
[\lim_{x\to 0}\frac{\tan x}{x}]<br />
    Is it same as
     <br />
\lim_{x\to 0}[\frac{\tan x}{x}]<br />

    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 \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 \lim_{x\to 0}[\frac{\sin x}{x}] =[\lim_{x\to 0}\frac{\sin x}{x}]
    and
    is \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 0 < x < \frac{\pi}{2}. then if you apply mean value theorem for f(t)=\sin t on the interval [0,x], then you get 0 < \frac{\sin x}{x} < 1.

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

    that interval. thus \lim_{x\to0} \lfloor{\frac{\sin x}{x} \rfloor}= \lim_{x\to 0} 0 = 0. but \lfloor{\lim_{x\to0} \frac{\sin x}{x} \rfloor}=\lfloor{1 \rfloor}=1. a similar argument works for \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
     <br />
[\lim_{x\to 0}\frac{\sin x}{x}]<br />
    we are required to first find out
     <br />
\lim_{x\to 0}\frac{\sin x}{x}<br />
    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 0<\frac{\sin x}{x}<1 the greatest integer should anyway be 0 whether we find
     <br />
[\lim_{x\to 0}\frac{\sin x}{x}]<br />
    or we find
     <br />
\lim_{x\to 0}[\frac{\sin x}{x}]<br />

    I am perplexed because in case of \frac{\tan x}{x} the answer for
     <br />
[\lim_{x\to 0}\frac{\tan x}{x}]=1<br />

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

     <br />
\lim_{x\to 0}[\frac{\tan x}{x}]=1<br />
    there's this general fact: suppose \lim_{x\to a}f(x)=b, and g is right continuous at x=b. if f(x) \geq b, in some neighbourhood of x=a, then \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 f(x) \geq b with f(x) \leq b. i'll prove the claim for the right continuous case: let \epsilon > 0. then since g is right continous at b,

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

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

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


    as an example you can check your question: the function g(x)=\lfloor{x \rfloor} is right continuous at x = 0 and 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 g to be the ceiling function, which is left continuous at x = 0, then you'll get the same result but for the function f(x)=\frac{\sin x}{x}.
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