Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

Commented by fotosy2k last updated on 27/Aug/21

pls belp

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

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

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

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com