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Question Number 88985 by jagoll last updated on 14/Apr/20

lim_(x→3) ln ∣x−3∣ exist or no?

$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{ln}\:\mid{x}−\mathrm{3}\mid\: \\ $$$${exist}\:{or}\:{no}? \\ $$

Commented by mr W last updated on 14/Apr/20

no.

$${no}. \\ $$

Commented by jagoll last updated on 14/Apr/20

because lim_(x→3) ln ∣x−3∣ = −∞ sir

$${because}\: \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{ln}\:\mid{x}−\mathrm{3}\mid\:=\:−\infty\:{sir} \\ $$

Commented by mr W last updated on 14/Apr/20

yes

$${yes} \\ $$

Commented by mr W last updated on 15/Apr/20

−∞ or +∞ is not a value! when you write lim_(x→a) f(x)=∞, it means in daily language that there is no limit. because the function can′t go close to a value. this is the same as when you say that you still have money even when the numver of money you have is zero.

$$−\infty\:{or}\:+\infty\:{is}\:{not}\:{a}\:{value}!\:{when}\:{you} \\ $$$${write}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)=\infty,\:{it}\:{means}\:{in}\:{daily} \\ $$$${language}\:{that}\:{there}\:{is}\:{no}\:{limit}.\:\:{because} \\ $$$${the}\:{function}\:{can}'{t}\:{go}\:{close}\:{to}\:{a}\:{value}. \\ $$$$ \\ $$$${this}\:{is}\:{the}\:{same}\:{as}\:{when}\:{you}\:{say}\:{that} \\ $$$${you}\:{still}\:{have}\:{money}\:{even}\:{when}\:{the} \\ $$$${numver}\:{of}\:{money}\:{you}\:{have}\:{is}\:{zero}. \\ $$

Commented by mr W last updated on 15/Apr/20

Commented by mr W last updated on 15/Apr/20

this is for the first day when lerning about limit.

$${this}\:{is}\:{for}\:{the}\:{first}\:{day}\:{when}\:{lerning} \\ $$$${about}\:{limit}. \\ $$

Commented by malwaan last updated on 15/Apr/20

lim_(x→3^+ ) ln∣x−3∣ = −∞ lim_(x→3^− ) ln∣x−3∣ = −∞ ⇒ lim_(x→3) ln∣x−3∣ = −∞ (so limit exist)or no ???

$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{3}^{+} } {\boldsymbol{{lim}}}\:\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{3}\mid\:=\:−\infty \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{3}^{−} } {\boldsymbol{{lim}}}\:\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{3}\mid\:=\:−\infty \\ $$$$\Rightarrow\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{3}} {\boldsymbol{{lim}}}\:\boldsymbol{{ln}}\mid\boldsymbol{{x}}−\mathrm{3}\mid\:=\:−\infty \\ $$$$\left({so}\:{limit}\:{exist}\right){or}\:{no}\:??? \\ $$

Commented by john santu last updated on 15/Apr/20

yes. i agree. lim_(x→3) ln ∣x−3∣ ⇒exist

$${yes}.\:{i}\:{agree}. \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{ln}\:\mid{x}−\mathrm{3}\mid\:\Rightarrow{exist}\: \\ $$

Commented by jagoll last updated on 15/Apr/20

i also think like that. by definition if lim_(x→a^+ ) f(x) = lim_(x→a^− ) f(x) then lim_(x→a) f(x) exist

$${i}\:{also}\:{think}\:{like}\:{that}. \\ $$$${by}\:{definition}\:{if}\: \\ $$$$\underset{{x}\rightarrow{a}^{+} } {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow{a}^{−} } {\mathrm{lim}}\:{f}\left({x}\right)\: \\ $$$${then}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{f}\left({x}\right)\:{exist}\: \\ $$

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