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

lim_(x→2) ((ax^2 +bx+6)/(x^2 −x−2))=10 ; find a=? ∧ b=?

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

Answered by deleteduser1 last updated on 03/Feb/24

((a(x^2 −x−2)+(b+a)x+6+2a)/(x^2 −x−2))=10 ⇒lim_(x→2) [a+(((b+a)x+6+2a)/((x−2)(x+1)))]=10 ⇒lim_(x→2) (((b+a)x+6+2a)/((x−2)(x+1)))=10−a 2(b+a)+6+2a=0⇒2a+b=−3 ...(i) ⇒lim_(x→2) (((b+a)x+6+2a)/((x−2)(x+1)))=lim_(x→2) ((b+a)/(2x−1))=((b+a)/3)=10−a ⇒b+a=30−3a⇒4a+b=30...(ii) (i)∧(ii)⇒a=16.5⇒b=−36

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

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

lim_(x→2) (ax^2 +bx+6)=a×2^2 +b×2+6=0 ⇒2a+b=−3 ..(i) lim_(x→2) =((ax^2 +bx+6)/(x^2 −x−2))=lim_(x→2) =((2ax+b)/(2x−1))=((2a×2+b)/(2×2−1))=10 ⇒4a+b=30 ...(ii) (ii)−(i): ⇒a=((33)/2)=16.5 ✓ ⇒b=−3−2×((33)/2)=−36 ✓

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

Commented by esmaeil last updated on 03/Feb/24

hi why do you use hopital?

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

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

why not? it′s there to be used.

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

Commented by esmaeil last updated on 03/Feb/24

ax^2 +bx+6]_(x=2) =^(??) 0

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

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

such that limit exists, ax^2 +bx+6]_(x=2) =^(must) 0. otherwise lim_(x→2) ((ax^2 +bx+6)/(x^2 −x−2))=∞

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

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...

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

Commented by esmaeil last updated on 03/Feb/24

ok. thanks. which is right? hopital or hospital.?

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

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

L′Hopital is right. but L′Hospital is also commonly spelled.

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

Commented by siyathokoza last updated on 07/Feb/24

lim_(x→2) (ax^2 +bx+6)=a×2^2 +b×2+6=0 ⇒2a+b=−3 ..(i) lim_(x→2) =((ax^2 +bx+6)/(x^2 −x−2))=lim_(x→2) =((2ax+b)/(2x−1))=((2a×2+b)/(2×2−1))=10 ⇒4a+b=30 ...(ii) (ii)−(i): ⇒a=((33)/2)=16.5 ✓ ⇒b=−3−2×((33)/2)=−36 ✓

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

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?

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

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

lim_(x→2) ((ax^2 +bx+6)/(x^2 −x−2))=10 ; find a=? ∧ b=? x^2 −x−2=(x−2)(x+1) ax^2 +bx+c=(x−2)(ax−3) =ax^2 −(3+2a)x+6 b=−(3+2a) lim_(x→2) ((ax^2 +bx+6)/(x^2 −x−2))=lim_(x→2) ((ax−3)/(x+1))=10 ((a(2)−3)/(2+1))=10 2a=33⇒a=33/2 b=−(3+2(33/2)=−36

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

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