Menu Close

Evaluate-i-lim-x-2-3x-3-4x-2-ii-lim-x-2-3x-3-4x-2-




Question Number 28876 by Rasheed.Sindhi last updated on 31/Jan/18
Evaluate  (i)  lim_(x→−∞)   ((2−3x)/( (√(3+4x^2 ))))  (ii)  lim_(x→+∞)   ((2−3x)/( (√(3+4x^2 ))))
$$\mathcal{E}{valuate} \\ $$$$\left(\mathrm{i}\right)\:\:\underset{{x}\rightarrow−\infty} {{lim}}\:\:\frac{\mathrm{2}−\mathrm{3x}}{\:\sqrt{\mathrm{3}+\mathrm{4x}^{\mathrm{2}} }} \\ $$$$\left(\mathrm{ii}\right)\:\:\underset{{x}\rightarrow+\infty} {{lim}}\:\:\frac{\mathrm{2}−\mathrm{3x}}{\:\sqrt{\mathrm{3}+\mathrm{4x}^{\mathrm{2}} }} \\ $$
Commented by abdo imad last updated on 31/Jan/18
i)=lim_(x→− ∞)    ((2−3x)/( (√(4x^2 (1+(3/(4x^2 )))))))=lim_(x→−∞)  ((−3x+2)/(2∣x∣(√(1+(3/(4x^2 ))))))  =lim_(x→−∞ )    ((−3x+2)/(−2x)) = (3/2)  ii)=lim_(x→+∞)  ((−3x+2)/(2∣x∣(√(1+(3/x^2 )))))=lim_(x→+∞)   ((−3x+2)/(2x))=−(3/2) .
$$\left.{i}\right)={lim}_{{x}\rightarrow−\:\infty} \:\:\:\frac{\mathrm{2}−\mathrm{3}{x}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}{x}^{\mathrm{2}} }\right)}}={lim}_{{x}\rightarrow−\infty} \:\frac{−\mathrm{3}{x}+\mathrm{2}}{\mathrm{2}\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}{x}^{\mathrm{2}} }}} \\ $$$$={lim}_{{x}\rightarrow−\infty\:} \:\:\:\frac{−\mathrm{3}{x}+\mathrm{2}}{−\mathrm{2}{x}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left.{ii}\right)={lim}_{{x}\rightarrow+\infty} \:\frac{−\mathrm{3}{x}+\mathrm{2}}{\mathrm{2}\mid{x}\mid\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }}}={lim}_{{x}\rightarrow+\infty} \:\:\frac{−\mathrm{3}{x}+\mathrm{2}}{\mathrm{2}{x}}=−\frac{\mathrm{3}}{\mathrm{2}}\:. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *