Is $\sin(x)/x$ continuous at $x=0$,(perfectly, not tends to)? The function has right hand and left hand limit same and equals 1 but what happens at perfect zero?it should be undefined. Is it?
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$\frac{\sin x}{x}$ is not defined at the origin. It is $$ \text{sinc}(x)=\left{\begin{array}{rcl}\frac{\sin x}{x}&\text{if}&x\neq 0 \ 1 & \text{if} & x=0 \end{array}\right.$$ the continuous (and much more: entire) function. – Jack D'Aurizio Jul 18 '17 at 17:08
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1For some basic information about writing math at this site see e.g. here, here, here and here. – Simply Beautiful Art Jul 18 '17 at 17:12
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The function $f(x)=\frac{\sin x}{x}$ is not defined at $x=0$. However, its limit is $1$ as $x\to 0$, so we can define a continuous function:
$$g(x)=\cases{\frac{\sin x}{x} & $x\ne 0$\\ 1 & $x=0$}$$
Since the value of $g(0)$ is equal to $\displaystyle\lim_{x\to 0}g(x)$, then we have that $g$ is continuous at $0$.
Simply Beautiful Art
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G Tony Jacobs
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It is undefined till you define it since the expression $\sin(x)/x$ is indeterminate at $x=0$. However it has a limit of $1$ as $x$ approaches zero. If you define a function to take the value $1$ when $x=0$ and the value $\sin(x)/x$ for $x\ne0$ then that function is continuous at zero.
spaceisdarkgreen
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You want to use the function f(x)=sin(x)/x and you find out that you can compute all its values, except for x=0, where you get 0/0, and that people call this "f(x)=sinx/x is not defined at x=0." But then, you also learn that smart Mathematicians have managed to find a solution by introducing the concept of limit: lim(x-->0){sinx/x}=1. You take your hat off fo the genius of Math, to make sure you compute the limit from the left and the limit from the right and find them equal 1 and so, now, you may think that you can plot the continuous function f(x)=sin(x)/x for all values, including x=0. However, for some reason, people still tell you that sin(x)/x is not defined at x=0. IMHO, this simply kills the smartness, applicability, and beauty of Math. Why is it not defined after people solved what previously "had no solution?" Some want you to define the function as
f(x)=sin(x)/x for x not 0 and f(x)=1 at x=0.
Why? Because some may want to define the function f(x)=sin(x)/x for x not 0 and f(x)=1217 at x=0 ??? They could, but why should they when the definition is simply sin(x)/x for all x and the previously uncertain value at x=0 has been solved long ago by the smart idea of limits?
Would you not be allowed to call the function sin(x) continuous because some may want to define it f(x)=sin(x) for all x but x=3 and f(x)=17 for x=3 ???
For the very important concept of derivative, you first compute the ratio (f(x)-f(x0))/(x-x0) when x is not x0, because otherwise, you get the same "not defined" 0/0. Then, you are taught that you get the result by using the same limit idea. If the claims on sin(x)/x are right, then the derivative of any function f(x) at any point x0 is not defined at x0. ((???)
