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Question Number 50089 by Cheyboy last updated on 13/Dec/18

$$\underset{x\to 8}{\lim }\frac{{\mathrm{x}}^2-6\mathrm{x}-16}{\mid \mathrm{x}-8\mid }$$

Commented by Abdo msup. last updated on 14/Dec/18

$$letf\left(x\right)=\frac{x^2-6x-16}{\mid x-8\mid }\Rightarrow f\left(x\right)=\frac{x^2-8x+2x-16}{\mid x-8\mid }$$$$=\frac{x(x-8)+2(x-8)}{\mid x-8\mid }=\frac{x-8}{\mid x-8\mid }(x+2)so$$$${lim}_{x\to 8^{+}}f\left(x\right)=10and{lim}_{x\to 8^{-}}=-10$$$$thoselimitsaredifferentssofdonthavelimitat{x}_{0^{}}=8$$$$butlimitsatleftandwrigtexist.$$

Commented by Cheyboy last updated on 14/Dec/18

$$\mathrm{Thank}\mathrm{you}\mathrm{guyz}$$

Answered by ajfour last updated on 13/Dec/18

$$ifx=8+h,(h>0)$$$$\underset{x\to 8}{\lim }\frac{(x-8)(x+2)}{\mid x-8\mid }=\underset{h\to 0}{\lim }\frac{h(10+h)}{h}=10$$$$ifx=8-h$$$$\underset{x\to 8}{\lim }\frac{(x-8)(x+2)}{\mid x-8\mid }=\underset{h\to 0}{\lim }\frac{-h(10-h)}{h}$$$$=-10$$$$\Rightarrow limit{doesn}^{\prime }texist.$$

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