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Question Number 4363 by Rasheed Soomro last updated on 13/Jan/16

If lim_(x→a) ((f(x))/(g(x))) exists and lim_(x→a) g(x)=0, then show that lim_(x→a) f(x)=0.

$${If}\:\:\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\:{exists}\:{and}\:\underset{{x}\rightarrow{a}} {{lim}}\:{g}\left({x}\right)=\mathrm{0}, \\ $$$${then}\:{show}\:{that} \\ $$$$\underset{{x}\rightarrow{a}} {{lim}}\:{f}\left({x}\right)=\mathrm{0}. \\ $$

Answered by Yozzii last updated on 13/Jan/16

Let l=lim_(x→a) ((f(x))/(g(x))) and h(x)=((f(x))/(g(x))) (g(x)≠0). l exists if, for every ε>0 there exists a δ>0 such that ∣h(x)−l∣<ε whenever 0<∣x−a∣<δ. ∴∣((f(x))/(g(x)))−lim_(x→a) ((f(x))/(g(x)))∣<ε ∣((f(x))/(g(x)))−((lim_(x→a) f(x))/(lim_(x→a) g(x)))∣<ε ∣((f(x)×lim_(x→a) g(x)−g(x)lim_(x→a) f(x))/(g(x)lim_(x→a) g(x)))∣<ε ∴∣f(x)lim_(x→a) g(x)−g(x)lim_(x→a) f(x)∣<ε∣g(x)lim_(x→a) g(x)∣ Since lim_(x→a) g(x)=0 we have ∣0−g(x)lim_(x→a) f(x)∣<0 Hence ∣−g(x)lim_(x→a) f(x)∣<0 which is false. (If indeed lim_(x→a) f(x)=0, there exists some error in my use of the ε−δ definition of the limit.)

$${Let}\:{l}=\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\:{and}\:{h}\left({x}\right)=\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\:\left({g}\left({x}\right)\neq\mathrm{0}\right).\:{l}\:{exists}\:{if},\:{for}\:{every} \\ $$$$\epsilon>\mathrm{0}\:{there}\:{exists}\:{a}\:\delta>\mathrm{0}\:{such}\:{that} \\ $$$$\mid{h}\left({x}\right)−{l}\mid<\epsilon\:{whenever}\:\mathrm{0}<\mid{x}−{a}\mid<\delta. \\ $$$$\therefore\mid\frac{{f}\left({x}\right)}{{g}\left({x}\right)}−\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\mid<\epsilon \\ $$$$\mid\frac{{f}\left({x}\right)}{{g}\left({x}\right)}−\frac{\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)}{\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)}\mid<\epsilon \\ $$$$\mid\frac{{f}\left({x}\right)×\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)−{g}\left({x}\right)\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)}{{g}\left({x}\right)\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)}\mid<\epsilon \\ $$$$\therefore\mid{f}\left({x}\right)\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)−{g}\left({x}\right)\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)\mid<\epsilon\mid{g}\left({x}\right)\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)\mid \\ $$$${Since}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)=\mathrm{0}\:{we}\:{have} \\ $$$$\mid\mathrm{0}−{g}\left({x}\right)\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)\mid<\mathrm{0} \\ $$$${Hence}\:\mid−{g}\left({x}\right)\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)\mid<\mathrm{0}\:{which}\:{is}\:{false}. \\ $$$$ \\ $$$$\left({If}\:{indeed}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0},\:{there}\:{exists}\right. \\ $$$${some}\:{error}\:{in}\:{my}\:{use}\:{of}\:{the}\:\epsilon−\delta\: \\ $$$$\left.{definition}\:{of}\:{the}\:{limit}.\right) \\ $$

Commented by prakash jain last updated on 13/Jan/16

lim_(x→3) ((x^2 −9)/(x−3))

$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{9}}{{x}−\mathrm{3}} \\ $$

Commented by Yozzii last updated on 13/Jan/16

My error was incorrectly using the result that lim_(x→a) ((f(x))/(g(x)))=((lim_(x→a) f(x))/(lim_(x→a) g(x))) given that lim_(x→a) g(x)≠0.

$${My}\:{error}\:{was}\:{incorrectly}\:{using}\:{the}\: \\ $$$${result}\:{that}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{f}\left({x}\right)}{{g}\left({x}\right)}=\frac{\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)}{\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)}\:{given} \\ $$$${that}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)\neq\mathrm{0}. \\ $$

Answered by RasheedSindhi last updated on 14/Jan/16

A way which doesn′t use ε-δ definition. Let lim_(x→a) ((f(x))/(g(x)))=l f(x)=((f(x))/(g(x)))×g(x) Applying limit to both sides lim_(x→a) f(x)=lim_(x→a) (((f(x))/(g(x)))×g(x)) =lim_(x→a) (((f(x))/(g(x)))×g(x))=lim_(x→a) ((f(x))/(g(x)))×lim_(x→a) g(x) =l.0=0

$${A}\:{way}\:{which}\:{doesn}'{t}\:{use}\:\epsilon-\delta \\ $$$${definition}. \\ $$$${Let}\:\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}={l} \\ $$$${f}\left({x}\right)=\frac{{f}\left({x}\right)}{{g}\left({x}\right)}×{g}\left({x}\right) \\ $$$${Applying}\:{limit}\:{to}\:{both}\:{sides} \\ $$$$\underset{{x}\rightarrow{a}} {{lim}f}\left({x}\right)=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{{f}\left({x}\right)}{{g}\left({x}\right)}×{g}\left({x}\right)\right) \\ $$$$=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{{f}\left({x}\right)}{{g}\left({x}\right)}×{g}\left({x}\right)\right)=\underset{{x}\rightarrow{a}} {{lim}}\frac{{f}\left({x}\right)}{{g}\left({x}\right)}×\underset{{x}\rightarrow{a}} {{lim}}\:{g}\left({x}\right) \\ $$$$={l}.\mathrm{0}=\mathrm{0} \\ $$

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