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Question Number 152431 by fotosy2k last updated on 28/Aug/21

Commented by fotosy2k last updated on 28/Aug/21

$$helppls$$

Commented by EDWIN88 last updated on 28/Aug/21

$$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(\frac{e^{5x}-e^{2x}}{x}\right)$$$$=\underset{x\to 0^{-}}{\lim }(\frac{e^{5x}-1}{x}+\frac{1-e^{2x}}{x})$$$$=\underset{x\to 0^{-}}{\lim }\left(\frac{e^{5x}-1}{x}\right)-\underset{x\to 0^{-}}{\lim }\left(\frac{e^{2x}-1}{x}\right)$$$$=5-2=3$$$$$$$$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\left(3\right)=3$$$$so\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)=3$$$$itfollowsthatf\left(x\right)continuatx=0$$

Commented by fotosy2k last updated on 30/Aug/21

$$thankyou$$

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