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Question Number 207943 by hardmath last updated on 31/May/24

$$\mathrm{Find}:$$$$\underset{\mathbf{x}\to 2^{-}}{\lim }\frac{(\mathrm{x}+2)\cdot (\mathrm{x}+1)}{\mid \mathrm{x}+2\mid }=?$$

Commented by mr W last updated on 31/May/24

$$=\frac{(2+2)(2+1)}{\mid 2+2\mid }=3$$

Commented by mr W last updated on 01/Jun/24

$$f\left(2\right)exists.so\underset{x\to 2}{\lim }f\left(x\right)=f\left(2\right).$$

Commented by hardmath last updated on 01/Jun/24

$$$$Dear professor, could you please explain

Commented by mr W last updated on 01/Jun/24

$$maybeyouwantedtoask$$$$\underset{\mathbf{x}\to -2^{-}}{\lim }\frac{(\mathrm{x}+2)\cdot (\mathrm{x}+1)}{\mid \mathrm{x}+2\mid }=?$$

Commented by hardmath last updated on 01/Jun/24

$$\mathrm{Dear}\mathrm{professor}...$$$$\underset{\mathbf{x}\to 2^{-}}{\lim }\frac{{\mathrm{x}}^4-16}{\mid \mathrm{x}-2\mid }=?$$

Commented by hardmath last updated on 01/Jun/24

$$$$Right limit and left limit

Commented by mr W last updated on 01/Jun/24

$$x\to 2^{-}meanx<2,i.e.\mid x-2\mid =2-x$$$$\underset{x\to 2^{-}}{\lim }\frac{x^4-16}{\mid x-2\mid }$$$$=\underset{x\to 2^{-}}{\lim }\frac{(x-2)(x+2)(x^2+4)}{2-x}$$$$=\underset{x\to 2^{-}}{\lim }\frac{(x+2)(x^2+4)}{-1}$$$$=\frac{(2+2)(2^2+4)}{-1}=-32$$

Commented by hardmath last updated on 01/Jun/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

Commented by hardmath last updated on 01/Jun/24

$$\mathrm{but}\mathrm{dear}\mathrm{pfofessor}$$$$$$Why not -x-2 from the module for x<2 in the above exercise

Commented by mr W last updated on 01/Jun/24

$$\mid x-2\mid =x-2ifx>2$$$$\mid x-2\mid =-(x-2)=2-xifx<2$$

Commented by hardmath last updated on 02/Jun/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

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