Mathematics - Limits Question with Solution | TestHub

A) $\underset{x\to 0+}{\mathrm{Lt}}x\left(\frac{e^{|x|+[x]}-2}{|x|+[x]}\right)=\underset{x\to 0+}{\mathrm{Lt}}x\left(\frac{e^{x+0}-2}{x+0}\right)=\underset{x\to 0}{\mathrm{Lt}}\left(e^{-x}-2\right)=1-2=-1$

B) $\underset{x\to 0-}{\mathrm{Lt}}x\left(\frac{e^{|x|+[x]}-2}{|x|+[x]}\right)=\underset{x\to 0-}{\mathrm{Lt}}x\left(\frac{e^{-x-1}-2}{-x-1}\right)=0$