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Question Number 155812 by zainaltanjung last updated on 05/Oct/21
Use algebraic simplifications to help find the limits, if they exist. 1). lim_(x→2) ((x^2 −4)/(x−2)) 2). lim_(x→1) ((x^2 −x)/(2x^2 +5x−7)) 3). lim_(x→1) ((3x^2 −13x−10)/(2x^2 −7x−15)) 4). lim_(x→1) ((3x^2 −13x−10)/(2x^2 −7x−15)) 5). lim_(x→2) ((x^3 −8)/(x^2 −4))
$$\mathrm{Use}\:\mathrm{algebraic}\:\mathrm{simplifications}\:\mathrm{to}\:\mathrm{help} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{limits},\:\mathrm{if}\:\mathrm{they}\:\mathrm{exist}. \\ $$$$\left.\mathrm{1}\right).\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{x}−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right).\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}−\mathrm{7}} \\ $$$$\left.\mathrm{3}\right).\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{13x}−\mathrm{10}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{7x}−\mathrm{15}} \\ $$$$\left.\mathrm{4}\right).\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{13x}−\mathrm{10}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{7x}−\mathrm{15}} \\ $$$$\left.\mathrm{5}\right).\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{8}}{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}\: \\ $$
Answered by Rasheed.Sindhi last updated on 05/Oct/21
1). lim_(x→2) ((x^2 −4)/(x−2)) =lim_(x→2) (((x−2)(x+2))/(x−2))=lim_(x→2) (x+2)=2+2=4 2). lim_(x→1) ((x^2 −x)/(2x^2 +5x−7))=lim_(x→1) ((x(x−1))/((x−1)(2x+7))) =lim_(x→1) (x/(2x+7))=(1/(2(1)+7))=(1/9) 3). lim_(x→1) ((3x^2 −13x−10)/(2x^2 −7x−15))=((3(1)^2 −13(1)−10)/(2(1)^2 −7(1)−15))=((−20)/(−20))=1 You can calculate limit directly because limit of denominator≠0 = lim_(x→1) ((3x^2 −13x−10)/(2x^2 −7x−15)) 4). lim_(x→1) ((3x^2 −13x−10)/(2x^2 −7x−15)) Same as Q3 5). lim_(x→2) ((x^3 −8)/(x^2 −4))=lim_(x→2) (((x−2)(x^2 +2x+4))/((x−2)(x+2))) =((2^2 +2.2+4)/(2+2))=((12)/4)=3
$$\left.\mathrm{1}\right).\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{x}−\mathrm{2}} \\ $$$$=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)}{{x}−\mathrm{2}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left({x}+\mathrm{2}\right)=\mathrm{2}+\mathrm{2}=\mathrm{4} \\ $$$$\left.\mathrm{2}\right).\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}−\mathrm{7}}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}\left({x}−\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{7}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}}{\mathrm{2}{x}+\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}\right)+\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\left.\mathrm{3}\right).\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{13x}−\mathrm{10}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{7x}−\mathrm{15}}=\frac{\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{13}\left(\mathrm{1}\right)−\mathrm{10}}{\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{7}\left(\mathrm{1}\right)−\mathrm{15}}=\frac{−\mathrm{20}}{−\mathrm{20}}=\mathrm{1} \\ $$$${You}\:{can}\:{calculate}\:{limit}\:{directly} \\ $$$${because}\:{limit}\:{of}\:{denominator}\neq\mathrm{0} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{13x}−\mathrm{10}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{7x}−\mathrm{15}} \\ $$$$\left.\mathrm{4}\right).\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{13x}−\mathrm{10}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{7x}−\mathrm{15}} \\ $$$$\:\:\:\:{Same}\:{as}\:{Q}\mathrm{3} \\ $$$$\left.\mathrm{5}\right).\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{8}}{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{2}.\mathrm{2}+\mathrm{4}}{\mathrm{2}+\mathrm{2}}=\frac{\mathrm{12}}{\mathrm{4}}=\mathrm{3} \\ $$
Commented by zainaltanjung last updated on 05/Oct/21
You are right but number 5 not yet true
$$\mathrm{You}\:\mathrm{are}\:\mathrm{right}\:\mathrm{but}\:\mathrm{number}\:\mathrm{5}\:\mathrm{not}\:\mathrm{yet}\:\mathrm{true} \\ $$
Commented by Rasheed.Sindhi last updated on 05/Oct/21
You can find these answers in a guide of the textbook.
$$\mathrm{You}\:\mathrm{can}\:\mathrm{find}\:\mathrm{these}\:\mathrm{answers}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{guide}\:\mathrm{of}\:\mathrm{the}\:\mathrm{textbook}. \\ $$
Commented by zainaltanjung last updated on 05/Oct/21
Okey...thank a lot
$$\mathrm{Okey}…\mathrm{thank}\:\mathrm{a}\:\mathrm{lot} \\ $$
Commented by puissant last updated on 05/Oct/21
Mr Sindhi x^3 −2=(x−2)(x^2 +2x+4).
$${Mr}\:{Sindhi}\:{x}^{\mathrm{3}} −\mathrm{2}=\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right). \\ $$
Commented by Rasheed.Sindhi last updated on 05/Oct/21
Yes sir you′re right.I have corrected.
$${Yes}\:{sir}\:{you}'{re}\:{right}.{I}\:{have}\:{corrected}. \\ $$
Commented by puissant last updated on 05/Oct/21
Thanks
$${Thanks} \\ $$
Answered by puissant last updated on 05/Oct/21
2) lim_(x→1) ((x^2 −x)/(2x^2 +5x−7))=lim_(x→1) ((x(x−1))/((x−1)(2x+7))) =lim_(x→1) (x/(2x+7)) = (1/9).. 3) lim_(x→1) ((3x^2 −13x−10)/(2x^2 −7x−15))= ((3−13−10)/(2−7−15))=((−20)/(−20))=1.. 4) = 1 5) lim_(x→2) ((x^3 −8)/(x^2 −4))=lim_(x→2) (((x−2)(x^2 +2x+4))/((x−2)(x+2))) =lim_(x→2) ((x^2 +2x+4)/(x+2))=((4+4+4)/4)=3..
$$\left.\mathrm{2}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} −{x}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{7}}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}\left({x}−\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{7}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}}{\mathrm{2}{x}+\mathrm{7}}\:=\:\frac{\mathrm{1}}{\mathrm{9}}.. \\ $$$$\left.\mathrm{3}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{13}{x}−\mathrm{10}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}−\mathrm{15}}=\:\frac{\mathrm{3}−\mathrm{13}−\mathrm{10}}{\mathrm{2}−\mathrm{7}−\mathrm{15}}=\frac{−\mathrm{20}}{−\mathrm{20}}=\mathrm{1}.. \\ $$$$\left.\mathrm{4}\right) \\ $$$$=\:\mathrm{1} \\ $$$$\left.\mathrm{5}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{\mathrm{3}} −\mathrm{8}}{{x}^{\mathrm{2}} −\mathrm{4}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}{{x}+\mathrm{2}}=\frac{\mathrm{4}+\mathrm{4}+\mathrm{4}}{\mathrm{4}}=\mathrm{3}.. \\ $$
Commented by zainaltanjung last updated on 05/Oct/21
okeyMr ....all of your answered is true
$$\mathrm{okeyMr}\:….\mathrm{all}\:\mathrm{of}\:\mathrm{your}\:\mathrm{answered}\:\mathrm{is}\:\mathrm{true}\: \\ $$

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