- #1

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I would like to understand the highlighted part. In my understanding, this function does not seem to have a hole! Having said this, i can state that ##x_0=1## and we have our defined ##f(x_0)=2##. It follows that,

##f(1^{+}) = e##

##f(1^{-}) = e##

thus ##f(x_0^{+})=f(x_0^{-})≠f(x_0)## thus the function has a removable discontinuity as per definition.