Suppose we are having two functions $$f(x )=\frac{\sin x(4-x^2)}{4x-x^3}$$ and $$g(x)=\frac{\sin x}{x}$$ they are not equal everywhere(for all real numbers), but they both are equal only in ${R}-${$0,2,-2$}

so while finding continuity writing the function $\frac{4-x^2}{4x-x^3}$ as $\frac{1}{x}$ is wrong.

but when we are evaluating the limit why do we take them equal? for example $$\lim_{x\to 0}{\frac{\sin x(4-x^2)}{4x-x^3}}$$ we write it as $$\lim_{x\to 0}{\frac{\sin x}{x}} $$ this means $$\frac{\sin x(4-x^2)}{4x-x^3}=\frac{\sin x}{x}$$ and this equality is totally wrong because as i have written above $f(x)\ne g(x)$, then while evaluating limit why do i take these two to be equal, this means we are changing our functions while evaluating the limit, i mean when we have to find $\lim_{x\to 0}f(x)$ we convert our function and find $\lim_{x\to 0}g(x)$.

And how our limit is correct if i changed my function.

I know limit means:

"How does a function behave when it is close to some number"

but when we evaluate the limit we are changing our function(from $f(x)$ to $g(x)$) then how can i expect that my limit of $f(x)$ is correct because i changed it to limit of $g(x)$

if i have misunderstood the meaning of limit, please do explain it to me :-)

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    $\begingroup$ Please could you try to re-write this question so it's better explained. $\endgroup$ – Fly by Night Apr 1 '16 at 15:54
  • $\begingroup$ ok i will edit :-) $\endgroup$ – ramsay Apr 1 '16 at 15:56
  • $\begingroup$ Writing $\frac{4-x^2}{4x-x^3}$ as $\frac{1}{x}$ is only wrong when $x = \pm 2$ as the first is undefined and the latter is well-defined. $\endgroup$ – Hetebrij Apr 1 '16 at 15:58
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    $\begingroup$ Maybe try separating the parts with $\sin$ into a different question... $\endgroup$ – T.J. Gaffney Apr 1 '16 at 15:58
  • $\begingroup$ wait i am editting, thanks for comments $\endgroup$ – ramsay Apr 1 '16 at 15:59

$$f(x)=\frac{\sin x(4-x^2)}{4x-x^3}=\frac{\sin x(4-x^2)}{x(4-x^2)} = \frac{\sin x}{x}, |x|\ne2$$

The function above is equal to $g(x)=\frac{\sin x}{x}$ everywhere but at the discontinuities $x=\pm2$. We see that through algebraic simplification, they have a very similar form. The major difference is in the domain of the functions: the former is not defined at $x=0,\pm2$, the latter is not defined at $x=0$.

When evaluating the limit, you've only shown interest in approaching $x\rightarrow 0$. Well, the two functions are identical in that locality, so they have the same behavior. You may ask then, what happens near another interesting point? What about $x\rightarrow \pm2$? Would my functions be different there? The answer in this case is yes, your functions would the different: we showed that earlier with the domains. However, their behavior in the limit is still the same for this example. Why? The discontinuity was removable!


We can do the work in a little more detail by using the fact that if $\lim_{x\to a} g(x)$ and $\lim_{x\to a} h(x)$ both exist, then $\lim_{x\to a} [g(x)h(x)]$ exists and $$ \lim_{x\to a} [g(x)h(x)] = \left[\lim_{x\to a} g(x)\right] \left[\lim_{x\to a} h(x)\right]. $$

For this exercise, we can let \begin{align} g(x) & = \frac{\sin x}{x}, \\ h(x) & = \frac{4-x^2}{4-x^2}. \end{align}

Then $f(x) = \frac{\sin x(4-x^2)}{4x-x^3} = g(x)h(x)$, so $$\lim_{x\to 0} f(x) = \lim_{x\to 0} [g(x)h(x)].$$

Now clearly $\lim_{x\to 0} h(x) = h(0) = 1$, so if $\lim_{x\to 0} g(x)$ exists then $$\lim_{x\to 0} [g(x)h(x)] = \lim_{x\to 0} g(x).$$

Alternatively, you can recognize that in a suitable neighborhood of zero, such as the set of all $x$ such that $0 < x < 1$ (for example), $f(x)$ and $g(x)$ are equal. Since the limit of each function as $x$ approaches zero is determined completely by the values of that function in some neighborhood of zero, the limits of the two functions must be equal. In other words, zahbaz's answer is correct.


This is a fundamental part of the concept of a limit. A limit of a function $f$ defined in a certain neighborhood of $a$ (with possible exception of the point $a$ itself) is defined in terms of values of the function near point $a$ (but not in terms of its value at $a$, in fact we don't even require $f$ to be defined at $a$). Hence if there are two functions $f$ and $g$ which are defined in a certain neighborhood of $a$ (except possibly at $a$) and $f(x) = g(x)$ for all $x$ in this neighborhood (except $x = a$) then the limiting behavior of both $f$ and $g$ as $x \to a$ is same. In your case the function $f$ and $g$ are such that $f(x) = g(x)$ if $x \in (-1, 1)$ and $x \neq 0$. Hence their limit is same as $x \to 0$.

I really don't understand why students have a hard time trying to understand this most often repeated fact that a limit as $x \to a$ is all about the action near $a$ but is totally independent of whatever action happens at $a$.


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