lim-x-2-ax-2-bx-6-x-2-x-2-10-find-a-b-

Question Number 203980 by Davidtim last updated on 03/Feb/24

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{ax}^{\mathrm{2}} +{bx}+\mathrm{6}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\mathrm{10}\:\:;\:\:\:{find}\:\:{a}=?\:\wedge\:{b}=? \\ $$

Answered by AST last updated on 03/Feb/24

$$\frac{{a}\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right)+\left({b}+{a}\right){x}+\mathrm{6}+\mathrm{2}{a}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\mathrm{10} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{2}} {{lim}}\left[{a}+\frac{\left({b}+{a}\right){x}+\mathrm{6}+\mathrm{2}{a}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}\right]=\mathrm{10} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\left({b}+{a}\right){x}+\mathrm{6}+\mathrm{2}{a}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}=\mathrm{10}−{a} \\ $$$$\mathrm{2}\left({b}+{a}\right)+\mathrm{6}+\mathrm{2}{a}=\mathrm{0}\Rightarrow\mathrm{2}{a}+{b}=−\mathrm{3}\:…\left({i}\right) \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\left({b}+{a}\right){x}+\mathrm{6}+\mathrm{2}{a}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}=\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{b}+{a}}{\mathrm{2}{x}−\mathrm{1}}=\frac{{b}+{a}}{\mathrm{3}}=\mathrm{10}−{a} \\ $$$$\Rightarrow{b}+{a}=\mathrm{30}−\mathrm{3}{a}\Rightarrow\mathrm{4}{a}+{b}=\mathrm{30}…\left({ii}\right) \\ $$$$\left({i}\right)\wedge\left({ii}\right)\Rightarrow{a}=\mathrm{16}.\mathrm{5}\Rightarrow{b}=−\mathrm{36} \\ $$

Answered by mr W last updated on 03/Feb/24

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left({ax}^{\mathrm{2}} +{bx}+\mathrm{6}\right)={a}×\mathrm{2}^{\mathrm{2}} +{b}×\mathrm{2}+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{a}+{b}=−\mathrm{3}\:\:\:..\left({i}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}=\frac{{ax}^{\mathrm{2}} +{bx}+\mathrm{6}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}=\frac{\mathrm{2}{ax}+{b}}{\mathrm{2}{x}−\mathrm{1}}=\frac{\mathrm{2}{a}×\mathrm{2}+{b}}{\mathrm{2}×\mathrm{2}−\mathrm{1}}=\mathrm{10} \\ $$$$\Rightarrow\mathrm{4}{a}+{b}=\mathrm{30}\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\Rightarrow{a}=\frac{\mathrm{33}}{\mathrm{2}}=\mathrm{16}.\mathrm{5}\:\checkmark \\ $$$$\Rightarrow{b}=−\mathrm{3}−\mathrm{2}×\frac{\mathrm{33}}{\mathrm{2}}=−\mathrm{36}\:\:\checkmark \\ $$

Commented by esmaeil last updated on 03/Feb/24

$${hi} \\ $$$${why}\:{do}\:\:{you}\:{use}\:{hopital}? \\ $$

Commented by mr W last updated on 03/Feb/24

$${why}\:{not}?\:{it}'{s}\:{there}\:{to}\:{be}\:{used}. \\ $$

Commented by esmaeil last updated on 03/Feb/24

$$\left.{ax}^{\mathrm{2}} +{bx}+\mathrm{6}\right]_{{x}=\mathrm{2}} \overset{??} {=}\mathrm{0} \\ $$

Commented by mr W last updated on 03/Feb/24

$${such}\:{that}\:{limit}\:{exists}, \\ $$$$\left.{ax}^{\mathrm{2}} +{bx}+\mathrm{6}\right]_{{x}=\mathrm{2}} \overset{{must}} {=}\mathrm{0}.\:{otherwise} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{ax}^{\mathrm{2}} +{bx}+\mathrm{6}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\infty \\ $$

Commented by esmaeil last updated on 03/Feb/24

$$\:\underline{\underbrace{\lesseqgtr}} \\ $$

Commented by Mathspace last updated on 03/Feb/24

$${sir}\:{smael}\:{hospital}\:{is}\:{good}\:{in}\:{this}\:{case}… \\ $$

Commented by esmaeil last updated on 03/Feb/24

$${ok}. \\ $$$${thanks}. \\ $$$${which}\:\:{is}\:\:{right}? \\ $$$${hopital}\:{or}\:{hospital}.? \\ $$

Commented by mr W last updated on 03/Feb/24

$${L}'{Hopital}\:{is}\:{right}.\:{but}\:{L}'{Hospital} \\ $$$${is}\:{also}\:{commonly}\:{spelled}. \\ $$

Commented by siyathokoza last updated on 07/Feb/24

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left({ax}^{\mathrm{2}} +{bx}+\mathrm{6}\right)={a}×\mathrm{2}^{\mathrm{2}} +{b}×\mathrm{2}+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{a}+{b}=−\mathrm{3}\:\:\:..\left({i}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}=\frac{{ax}^{\mathrm{2}} +{bx}+\mathrm{6}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}=\frac{\mathrm{2}{ax}+{b}}{\mathrm{2}{x}−\mathrm{1}}=\frac{\mathrm{2}{a}×\mathrm{2}+{b}}{\mathrm{2}×\mathrm{2}−\mathrm{1}}=\mathrm{10} \\ $$$$\Rightarrow\mathrm{4}{a}+{b}=\mathrm{30}\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\Rightarrow{a}=\frac{\mathrm{33}}{\mathrm{2}}=\mathrm{16}.\mathrm{5}\:\checkmark \\ $$$$\Rightarrow{b}=−\mathrm{3}−\mathrm{2}×\frac{\mathrm{33}}{\mathrm{2}}=−\mathrm{36}\:\:\checkmark \\ $$

Commented by mr W last updated on 07/Feb/24

$${does}\:{it}\:{make}\:{any}\:{sense}\:{that}\:{you}\: \\ $$$${post}\:{a}\:{copy}\:{from}\:{other}\:{people}'{s} \\ $$$${workings}? \\ $$

Answered by Rasheed.Sindhi last updated on 03/Feb/24

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{ax}^{\mathrm{2}} +{bx}+\mathrm{6}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\mathrm{10}\:\:;\:\:\:{find}\:\:{a}=?\:\wedge\:{b}=? \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{2}=\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right) \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\left({x}−\mathrm{2}\right)\left({ax}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={ax}^{\mathrm{2}} −\left(\mathrm{3}+\mathrm{2}{a}\right){x}+\mathrm{6} \\ $$$${b}=−\left(\mathrm{3}+\mathrm{2}{a}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{ax}^{\mathrm{2}} +{bx}+\mathrm{6}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{ax}−\mathrm{3}}{{x}+\mathrm{1}}=\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{a}\left(\mathrm{2}\right)−\mathrm{3}}{\mathrm{2}+\mathrm{1}}=\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{a}=\mathrm{33}\Rightarrow{a}=\mathrm{33}/\mathrm{2} \\ $$$$\:\:\:\:{b}=−\left(\mathrm{3}+\mathrm{2}\left(\mathrm{33}/\mathrm{2}\right)=−\mathrm{36}\right. \\ $$

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