Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 26289 by d.monhbayr@gmail.com last updated on 23/Dec/17

lim_(x→2) ((1/(x−2))−((12)/(x^3 −8)))

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}−\mathrm{2}}−\frac{\mathrm{12}}{{x}^{\mathrm{3}} −\mathrm{8}}\right) \\ $$

Answered by prakash jain last updated on 24/Dec/17

(1/(x−2))−(1/((x−2)(x^2 +2x+4))) =((x^2 +2x+4−12)/((x−2)(x^2 +2x+4))) =((x^2 +2x−8)/((x−2)(x^2 +2x+4)))=(((x−2)(x+4))/((x−2)(x^2 +2x+4))) lim_(x→2) (((x−2)(x+4))/((x−2)(x^2 +2x+4)))=((2+4)/(4+4+4))=(1/2)

$$\frac{\mathrm{1}}{{x}−\mathrm{2}}−\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}−\mathrm{12}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}=\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{4}\right)}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{4}\right)}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}=\frac{\mathrm{2}+\mathrm{4}}{\mathrm{4}+\mathrm{4}+\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Rasheed.Sindhi last updated on 24/Dec/17

pl check last step ((2+4)/(4+4+4))=(6/(12))=(1/2)

$$\mathrm{pl}\:\mathrm{check}\:\mathrm{last}\:\mathrm{step}\:\frac{\mathrm{2}+\mathrm{4}}{\mathrm{4}+\mathrm{4}+\mathrm{4}}=\frac{\mathrm{6}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 24/Dec/17

Thanks Corrected

Terms of Service

Privacy Policy

Contact: info@tinkutara.com