Continuity in Composite functions

**pankaj** · September 4th 2008, 04:35 AM

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

Another example that i have is:
$f(x)=[x]$
$ g(x)=\frac{tanx}{x} $
In this case is
$ \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

**ThePerfectHacker** · September 4th 2008, 06:49 AM

Originally Posted by pankaj

If $\lim_{x\to a}g(x)$ exists then is it proper to say that
$ \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 $\lim_{x\to a}g(x)=L$ 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, $\lim_{x\to a}f(g(x)) = f(L)$ .

**pankaj** · September 4th 2008, 08:29 AM

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
$ \lim_{x\to 0}[\frac{\sin x}{x}]=0 $

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

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

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

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

This is what I want to verify

**Plato** · September 4th 2008, 08:47 AM

Look at the graphs of the two functions.
Then think in therms of the floor function.
It should be clear that $\lim _{x \to 0} \left( {\left\lfloor {\frac{{\sin (x)}}{x}} \right\rfloor } \right) = 0\,\& \,\lim _{x \to 0} \left( {\left\lfloor {\frac{{\tan (x)}}{x}} \right\rfloor } \right) = 1$ .

**pankaj** · September 4th 2008, 09:05 AM

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

**Plato** · September 4th 2008, 09:16 AM

Well, have you looked at the graphs.

**pankaj** · September 4th 2008, 09:35 AM

I get your point

**NonCommAlg** · September 4th 2008, 10:03 AM

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 $\frac{\tan x}{x}.$

**pankaj** · September 4th 2008, 05:55 PM

This is precisely what I wanted to know that while evaluating
$ [\lim_{x\to 0}\frac{\sin x}{x}] $
we are required to first find out
$ \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 $0<\frac{\sin x}{x}<1$ the greatest integer should anyway be 0 whether we find
$ [\lim_{x\to 0}\frac{\sin x}{x}] $
or we find
$ \lim_{x\to 0}[\frac{\sin x}{x}] $

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

$ \lim_{x\to 0}[\frac{\tan x}{x}]=1 $

**NonCommAlg** · September 5th 2008, 01:44 AM

Originally Posted by pankaj

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

$ \lim_{x\to 0}[\frac{\tan x}{x}]=1 $

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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