1. 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
Also i must mention that [x] denotes the greatest integer less than or equal to x

2. Originally Posted by pankaj
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, $\displaystyle \lim_{x\to a}f(g(x)) = f(L)$.

3. 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

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

5. 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

6. Well, have you looked at the graphs.

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

9. 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
$\displaystyle [\lim_{x\to 0}\frac{\tan x}{x}]=1$

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

10. Originally Posted by pankaj
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 $\displaystyle f(x)=\frac{\sin x}{x}.$