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Question Number 76191 by abdomathmax last updated on 25/Dec/19
find lim_(x→0)     ((e^x −e^([x]) )/x)
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{e}^{{x}} −{e}^{\left[{x}\right]} }{{x}} \\ $$
Answered by Rio Michael last updated on 25/Dec/19
Am not sure if this limit  would exist because   lim_(x→0^−  )  ((e^x −e^([x]) )/x) = 1 as  e^([x]) → 0 as x→0  lim_(x→0^+  )  ((e^x −e^([x]) )/x) = 0
$${Am}\:{not}\:{sure}\:{if}\:{this}\:{limit} \\ $$$${would}\:{exist}\:{because}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} \:} {{lim}}\:\frac{{e}^{{x}} −{e}^{\left[{x}\right]} }{{x}}\:=\:\mathrm{1}\:{as}\:\:{e}^{\left[{x}\right]} \rightarrow\:\mathrm{0}\:{as}\:{x}\rightarrow\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} \:} {{lim}}\:\frac{{e}^{{x}} −{e}^{\left[{x}\right]} }{{x}}\:=\:\mathrm{0}\: \\ $$$$ \\ $$
Commented by MJS last updated on 25/Dec/19
lim_(x→0^− ) ((e^x −e^([x]) )/x) =−∞  −1≤x<0 ⇒ e^([x]) =(1/e)  ⇒ as x→0^−  we get ((1−(1/e))/(x→0^− ))=−∞
$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{\mathrm{e}^{{x}} −\mathrm{e}^{\left[{x}\right]} }{{x}}\:=−\infty \\ $$$$−\mathrm{1}\leqslant{x}<\mathrm{0}\:\Rightarrow\:\mathrm{e}^{\left[{x}\right]} =\frac{\mathrm{1}}{\mathrm{e}} \\ $$$$\Rightarrow\:\mathrm{as}\:{x}\rightarrow\mathrm{0}^{−} \:\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}}}{{x}\rightarrow\mathrm{0}^{−} }=−\infty \\ $$

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