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Question Number 227080 by Spillover last updated on 29/Dec/25

Answered by Kassista last updated on 29/Dec/25

a) lim_(x→−6) f(x) = lim_(x→−6) 7−4x = 7−4(−6)= 31 b) lim_(x→1^− ) f(x) = lim_(x→1^− ) 7−4x = 7−4(1)=3 and lim_(x→1^+ ) f(x) = lim_(x→1^+ ) x^2 +2x = 1+2(1)=3 since lim_(x→1−) f(x)=1=lim_(x→1^+ ) f(x) ⇒ lim_(x→1) f(x)=1

$$\left.{a}\right)\:\underset{{x}\rightarrow−\mathrm{6}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow−\mathrm{6}} {\mathrm{lim}}\:\mathrm{7}−\mathrm{4}{x}\:=\:\mathrm{7}−\mathrm{4}\left(−\mathrm{6}\right)=\:\mathrm{31} \\ $$$$ \\ $$$$\left.{b}\right)\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\mathrm{7}−\mathrm{4}{x}\:=\:\mathrm{7}−\mathrm{4}\left(\mathrm{1}\right)=\mathrm{3} \\ $$$$\:\:{and}\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:{x}^{\mathrm{2}} +\mathrm{2}{x}\:=\:\mathrm{1}+\mathrm{2}\left(\mathrm{1}\right)=\mathrm{3} \\ $$$${since}\:\underset{{x}\rightarrow\mathrm{1}−} {\mathrm{lim}}\:{f}\left({x}\right)=\mathrm{1}=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:{f}\left({x}\right)\:\Rightarrow\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{1} \\ $$$$ \\ $$

Commented by Spillover last updated on 11/Jan/26

thanks

$${thanks} \\ $$

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