Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 119253 by bemath last updated on 23/Oct/20

Given three function f(x) ,g(x) and h(x).  where f(x)=x^2 +x−2 and h(x)=x^2 +2x−1.  If ((f(x))/(x+3)) ≤ ((g(x))/(x+3)) ≤ ((h(x))/(x+3)) , then the value of  lim_(x→−1)  g(x) = ?

$${Given}\:{three}\:{function}\:{f}\left({x}\right)\:,{g}\left({x}\right)\:{and}\:{h}\left({x}\right). \\ $$$${where}\:{f}\left({x}\right)={x}^{\mathrm{2}} +{x}−\mathrm{2}\:{and}\:{h}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}. \\ $$$${If}\:\frac{{f}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\frac{{g}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\frac{{h}\left({x}\right)}{{x}+\mathrm{3}}\:,\:{then}\:{the}\:{value}\:{of} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:{g}\left({x}\right)\:=\:? \\ $$

Answered by benjo_mathlover last updated on 23/Oct/20

f(x)=(x+2)(x−1)  h(x)=(x+1)^2 −2  lim_(x→−1)  ((f(x))/(x+3)) ≤ lim_(x→−1)  ((g(x))/(x+3)) ≤ lim_(x→−1)  ((h(x))/(x+3))  −(1/2)lim_(x→−1)  f(x)≤ −(1/2)lim_(x→−1)  g(x) ≤−(1/2)lim_(x→−1) h(x)  ⇔ lim_(x→−1) h(x)≤ lim_(x→−1) g(x) ≤ lim_(x→−1) f(x)  where lim_(x→−1) h(x)=−2 ∧ lim_(x→−1) f(x)=−2  so we get lim_(x→−1) g(x)=−2

$${f}\left({x}\right)=\left({x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right) \\ $$$${h}\left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{{g}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{{h}\left({x}\right)}{{x}+\mathrm{3}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:{f}\left({x}\right)\leqslant\:−\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:{g}\left({x}\right)\:\leqslant−\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}{h}\left({x}\right) \\ $$$$\Leftrightarrow\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}{h}\left({x}\right)\leqslant\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}{g}\left({x}\right)\:\leqslant\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right) \\ $$$${where}\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}{h}\left({x}\right)=−\mathrm{2}\:\wedge\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=−\mathrm{2} \\ $$$${so}\:{we}\:{get}\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}{g}\left({x}\right)=−\mathrm{2} \\ $$

Answered by 1549442205PVT last updated on 23/Oct/20

when x→−1 then x+3>0.Hence   ((f(x))/(x+3)) ≤ ((g(x))/(x+3)) ≤ ((h(x))/(x+3))  ⇔f(x)≤g(x)≤h(x).We infer that  lim_(→−1) f(x)≤lim_(x→−1) g(x)≤lim_(x→−1) h(x)  ⇔lim_(x→−1) (x^2 +x−2)≤lim_(x→−1) g(x)≤lim_(x→−1) (x^2 +2x−1)  ⇔−2≤lim_(x→−1) g(x)≤−2  ⇒lim_(x→−1) g(x)=−2

$$\mathrm{when}\:\mathrm{x}\rightarrow−\mathrm{1}\:\mathrm{then}\:\mathrm{x}+\mathrm{3}>\mathrm{0}.\mathrm{Hence} \\ $$$$\:\frac{{f}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\frac{{g}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\frac{{h}\left({x}\right)}{{x}+\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{f}\left(\mathrm{x}\right)\leqslant\mathrm{g}\left(\mathrm{x}\right)\leqslant\mathrm{h}\left(\mathrm{x}\right).\mathrm{We}\:\mathrm{infer}\:\mathrm{that} \\ $$$$\underset{\rightarrow−\mathrm{1}} {\mathrm{lim}f}\left(\mathrm{x}\right)\leqslant\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{lim}g}\left(\mathrm{x}\right)\leqslant\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{lim}h}\left(\mathrm{x}\right) \\ $$$$\Leftrightarrow\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{2}\right)\leqslant\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{lim}g}\left(\mathrm{x}\right)\leqslant\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{1}\right) \\ $$$$\Leftrightarrow−\mathrm{2}\leqslant\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{lim}g}\left(\mathrm{x}\right)\leqslant−\mathrm{2} \\ $$$$\Rightarrow\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{lim}g}\left(\mathrm{x}\right)=−\mathrm{2} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com