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Question Number 155812 by zainaltanjung last updated on 05/Oct/21

$$\mathrm{Use}\mathrm{algebraic}\mathrm{simplifications}\mathrm{to}\mathrm{help}$$$$\mathrm{find}\mathrm{the}\mathrm{limits},\mathrm{if}\mathrm{they}\mathrm{exist}.$$$$1).\underset{x\to 2}{\lim }\frac{{\mathrm{x}}^2-4}{\mathrm{x}-2}$$$$2).\underset{x\to 1}{\lim }\frac{{\mathrm{x}}^2-\mathrm{x}}{{2\mathrm{x}}^2+5\mathrm{x}-7}$$$$3).\underset{x\to 1}{\lim }\frac{{3\mathrm{x}}^2-13\mathrm{x}-10}{{2\mathrm{x}}^2-7\mathrm{x}-15}$$$$4).\underset{x\to 1}{\lim }\frac{{3\mathrm{x}}^2-13\mathrm{x}-10}{{2\mathrm{x}}^2-7\mathrm{x}-15}$$$$5).\underset{x\to 2}{\lim }\frac{{\mathrm{x}}^3-8}{{\mathrm{x}}^2-4}$$

Answered by Rasheed.Sindhi last updated on 05/Oct/21

$$1).\underset{x\to 2}{\lim }\frac{{\mathrm{x}}^2-4}{\mathrm{x}-2}$$$$=\underset{x\to 2}{\lim }\frac{(x-2)(x+2)}{x-2}=\underset{x\to 2}{\lim }(x+2)=2+2=4$$$$2).\underset{x\to 1}{\lim }\frac{{\mathrm{x}}^2-\mathrm{x}}{{2\mathrm{x}}^2+5\mathrm{x}-7}=\underset{x\to 1}{\lim }\frac{x(x-1)}{(x-1)(2x+7)}$$$$=\underset{x\to 1}{\lim }\frac{x}{2x+7}=\frac{1}{2\left(1\right)+7}=\frac{1}{9}$$$$3).\underset{x\to 1}{\lim }\frac{{3\mathrm{x}}^2-13\mathrm{x}-10}{{2\mathrm{x}}^2-7\mathrm{x}-15}=\frac{3{\left(1\right)}^2-13\left(1\right)-10}{2{\left(1\right)}^2-7\left(1\right)-15}=\frac{-20}{-20}=1$$$$Youcancalculatelimitdirectly$$$$becauselimitofdenominator\ne 0$$$$=\underset{x\to 1}{\lim }\frac{{3\mathrm{x}}^2-13\mathrm{x}-10}{{2\mathrm{x}}^2-7\mathrm{x}-15}$$$$4).\underset{x\to 1}{\lim }\frac{{3\mathrm{x}}^2-13\mathrm{x}-10}{{2\mathrm{x}}^2-7\mathrm{x}-15}$$$$SameasQ3$$$$5).\underset{x\to 2}{\lim }\frac{{\mathrm{x}}^3-8}{{\mathrm{x}}^2-4}=\underset{x\to 2}{\lim }\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}$$$$=\frac{2^2+2.2+4}{2+2}=\frac{12}{4}=3$$

Commented by zainaltanjung last updated on 05/Oct/21

$$\mathrm{You}\mathrm{are}\mathrm{right}\mathrm{but}\mathrm{number}5\mathrm{not}\mathrm{yet}\mathrm{true}$$

Commented by Rasheed.Sindhi last updated on 05/Oct/21

$$\mathrm{You}\mathrm{can}\mathrm{find}\mathrm{these}\mathrm{answers}\mathrm{in}\mathrm{a}$$$$\mathrm{guide}\mathrm{of}\mathrm{the}\mathrm{textbook}.$$

Commented by zainaltanjung last updated on 05/Oct/21

$$\mathrm{Okey}...\mathrm{thank}\mathrm{a}\mathrm{lot}$$

Commented by puissant last updated on 05/Oct/21

$$MrSindhi{x}^3-2=(x-2)(x^2+2x+4).$$

Commented by Rasheed.Sindhi last updated on 05/Oct/21

$$Yessir{you}^{\prime }reright.Ihavecorrected.$$

Commented by puissant last updated on 05/Oct/21

$$Thanks$$

Answered by puissant last updated on 05/Oct/21

$$2)$$$$\underset{x\to 1}{\lim }\frac{x^2-x}{2x^2+5x-7}=\underset{x\to 1}{\lim }\frac{x(x-1)}{(x-1)(2x+7)}$$$$=\underset{x\to 1}{\lim }\frac{x}{2x+7}=\frac{1}{9}..$$$$3)$$$$\underset{x\to 1}{\lim }\frac{3x^2-13x-10}{2x^2-7x-15}=\frac{3-13-10}{2-7-15}=\frac{-20}{-20}=1..$$$$4)$$$$=1$$$$5)$$$$\underset{x\to 2}{\lim }\frac{x^3-8}{x^2-4}=\underset{x\to 2}{\lim }\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}$$$$=\underset{x\to 2}{\lim }\frac{x^2+2x+4}{x+2}=\frac{4+4+4}{4}=3..$$

Commented by zainaltanjung last updated on 05/Oct/21

$$\mathrm{okeyMr}....\mathrm{all}\mathrm{of}\mathrm{your}\mathrm{answered}\mathrm{is}\mathrm{true}$$

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