Question-152371

Question Number 152371 by fotosy2k last updated on 27/Aug/21

Commented by fotosy2k last updated on 27/Aug/21

$${pls}\:{belp} \\ $$$$ \\ $$

Answered by Olaf_Thorendsen last updated on 27/Aug/21

lim_(x→2) ((x−2)/(x^2 −3x+2)) = lim _(x→2) ((x−2)/((x−2)(x−1))) = lim_(x→2) (1/(x−1)) = 1

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}\:=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}\:}\frac{{x}−\mathrm{2}}{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}−\mathrm{1}}\:=\:\mathrm{1} \\ $$

Commented by fotosy2k last updated on 27/Aug/21

$${thank}\:{you} \\ $$

Answered by Rasheed.Sindhi last updated on 28/Aug/21

AnOther Approach lim_(x→2) ((x−2)/(x^2 −3x+2)) x=2+h⇒x→2⇒h→0 lim_(h→0) ((2+h−2)/((2+h)^2 −3(2+h)+2)) lim_(h→0) (h/(4+4h+h^2 −6−3h+2)) lim_(h→0) (h/(h+h^2 ))=lim_(h→h) (1/(1+h))=(1/(1+0))=1

$$\boldsymbol{\mathrm{AnOther}}\:\boldsymbol{\mathrm{Approach}} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}} \\ $$$${x}=\mathrm{2}+{h}\Rightarrow{x}\rightarrow\mathrm{2}\Rightarrow{h}\rightarrow\mathrm{0} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{2}+{h}−\mathrm{2}}{\left(\mathrm{2}+{h}\right)^{\mathrm{2}} −\mathrm{3}\left(\mathrm{2}+{h}\right)+\mathrm{2}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{h}}{\mathrm{4}+\mathrm{4}{h}+{h}^{\mathrm{2}} −\mathrm{6}−\mathrm{3}{h}+\mathrm{2}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{h}}{{h}+{h}^{\mathrm{2}} }=\underset{{h}\rightarrow{h}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+{h}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{0}}=\mathrm{1} \\ $$$$ \\ $$

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