Question-176710

Question Number 176710 by eka last updated on 25/Sep/22

Answered by a.lgnaoui last updated on 26/Sep/22

3. lim_(x→−1) ((√(x^2 +3))/(x^2 −2))=((√4)/(−1))=−2 anser (D) 4. ((x^2 +2x−8)/(x−2))=(((x−2)(x+4))/(x−2))=x+4 lim((x^2 +2x−8)/(x−2))=2+4=6 anser (D)

$$\:\mathrm{3}.\:\:\:{lim}_{{x}\rightarrow−\mathrm{1}} \:\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{{x}^{\mathrm{2}} −\mathrm{2}}=\frac{\sqrt{\mathrm{4}}}{−\mathrm{1}}=−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:{anser}\:\left(\boldsymbol{\mathrm{D}}\right) \\ $$$$\mathrm{4}.\:\:\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}}{{x}−\mathrm{2}}=\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{4}\right)}{{x}−\mathrm{2}}={x}+\mathrm{4} \\ $$$$\:\:\:{lim}\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}}{{x}−\mathrm{2}}=\mathrm{2}+\mathrm{4}=\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:{anser}\:\:\left(\boldsymbol{\mathrm{D}}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 25/Sep/22

1. lim_(x→2) ((4x+1)/(3−2x)) ∵ lim_(x→2) (3−2x)≠0 ∴ lim_(x→2) ((4x+1)/(3−2x))=((4(1)+1)/(3−2(1)))=(5/(−1))=−5 2. lim_(x→1) ((x^2 −3x−4)/(x^2 −2x−3))=((1^2 −3(1)−4)/((1)^2 −2(1)−3(≠0))) =((−6)/(−4))=(3/2)

$$\mathrm{1}.\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{4}{x}+\mathrm{1}}{\mathrm{3}−\mathrm{2}{x}} \\ $$$$\because\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left(\mathrm{3}−\mathrm{2}{x}\right)\neq\mathrm{0} \\ $$$$\therefore\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{4}{x}+\mathrm{1}}{\mathrm{3}−\mathrm{2}{x}}=\frac{\mathrm{4}\left(\mathrm{1}\right)+\mathrm{1}}{\mathrm{3}−\mathrm{2}\left(\mathrm{1}\right)}=\frac{\mathrm{5}}{−\mathrm{1}}=−\mathrm{5} \\ $$$$\mathrm{2}.\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}}=\frac{\mathrm{1}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{1}\right)−\mathrm{4}}{\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}\right)−\mathrm{3}\left(\neq\mathrm{0}\right)} \\ $$$$=\frac{−\mathrm{6}}{−\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Sep/22

3.lim_(x→−1) ((√(x^2 +3))/(x^2 −2))=((√((−1)^2 +3))/((−1)^2 −2 (≠0))) =(2/(−1))=−2 4.lim_(x→2) ((x^2 +2x−8)/(x−2))=lim_(x→2) (((x−2)(x+4))/(x−2)) =2+4=6

$$\mathrm{3}.\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{{x}^{\mathrm{2}} −\mathrm{2}}=\frac{\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}}}{\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\:\left(\neq\mathrm{0}\right)} \\ $$$$=\frac{\mathrm{2}}{−\mathrm{1}}=−\mathrm{2} \\ $$$$\mathrm{4}.\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}}{{x}−\mathrm{2}}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\left(\cancel{{x}−\mathrm{2}}\right)\left({x}+\mathrm{4}\right)}{\cancel{{x}−\mathrm{2}}} \\ $$$$=\mathrm{2}+\mathrm{4}=\mathrm{6} \\ $$

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