MHB Is the Discontinuity in $\frac{\sin x}{x}$ at $x=0$ a Jump Type?

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The function f(x) = sin(x)/x is defined as 0 at x=0, while the limit as x approaches 0 is 1. This creates a discontinuity at x=0. The reasoning presented suggests it is a jump discontinuity, but it is actually classified as a removable discontinuity. Thus, the discontinuity can be resolved by redefining f(0) to equal 1.
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$f(x) = \frac{\sin x}{x}$ if $x\neq 0$, $f(0) = 0$.We know that the $\lim\limits_{n\to 0}f(x) = 1$. Since $f(0) = 0$, we will have a jump discontinuity.
Is this correct reasoning?
 
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dwsmith said:
$f(x) = \frac{\sin x}{x}$ if $x\neq 0$, $f(0) = 0$.We know that the $\lim\limits_{n\to 0}f(x) = 1$. Since $f(0) = 0$, we will have a jump discontinuity.
Is this correct reasoning?

No this is a removable discontinuity. :)
 
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