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I have been told that you cannot derive a function at a removable discontinuity. The derivative is defined as: $$ \lim_{h\to0}\frac{f(x+h) - f(x)}{h} $$

And if there is a removable discontinuity at $a$ then you cannot input a value for $ f(a)$ and thus cannot calculate the derivative.

However, what if we use $ \lim_{x\to a}$ and input that value instead of $ f(a)$? Although the point is undefined, the limit still exists. Is there a problem with doing this? Are there equations where doing so doesn't make sense? Or was I right all along?

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    $\begingroup$ possible duplicate of Derivative on removable discontinuity $\endgroup$ Commented Sep 30, 2015 at 1:31
  • $\begingroup$ If you fill the hole at $x=a$ and want to calculate the derivative at $a$, then you cannot just take the derivative with the power rule, product rule etc. You are supposed to use the formal definition of the derivative as you stated. (Just filling the hole though a piece wise function, does not make the original function "algebraically "continuous) $\endgroup$
    – imranfat
    Commented Sep 30, 2015 at 1:31

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The geometrical definition of a derivative is the slope of the tangent line at a point. The slope does not exist at f(a) if it's undefined at that point, since slope is equal to (f(a)-f(b))/a-b

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