Lim_x→ ∞ ( √(4x^2 7x) +2x ) Substitute y= x,as x →∞,1/x→ 0 =lim_y→∞ ( √(4y^2 7y) 2y ) =lim_y→∞ ( √(4y^2+7y) 2y ) ( √(4y^2+7y) +2y )/ ( √(4y^2+7y) +2y ) =lim_ ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x →-∞√4 x...

Question

${lim}_{x \to - \infty} (\sqrt{4 x^{2} - 7 x} + 2 x)$

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Solution

${lim}_{x \to - \infty} (\sqrt{4 x^{2} - 7 x} + 2 x)$

Substitute y=-x,as $x \to \infty, \frac{1}{x} \to 0$

$= {lim}_{y \to \infty} (\sqrt{4 y^{2} - 7 y} - 2 y)$

$= {lim}_{y \to \infty} \frac{(\sqrt{4 y^{2} + 7 y} - 2 y) (\sqrt{4 y^{2} + 7 y} + 2 y)}{(\sqrt{4 y^{2} + 7 y} + 2 y)}$

$= {lim}_{y \to \infty} \frac{(4 y^{2} + 7 y - 4 y^{2})}{(\sqrt{4 y^{2} + 7 y} + 2 y)}$

$= {lim}_{y \to \infty} \frac{(7 y)}{(\sqrt{4 y^{2} - 7 y} + 2 y)}$

$= {lim}_{y \to \infty} \frac{7}{(\sqrt{4 + \frac{7}{y}} + 2)}$

$= \frac{7}{2 + 2} = \frac{7}{4}$

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limx→−∞(√4x2−7x+2x)

limx→−∞(√4x2−7x+2x) Substitute y=-x,as x→∞,1x→0 =limy→∞(√4y2−7y−2y) =limy→∞(√4y2+7y−2y)(√4y2+7y+2y)(√4y2+7y+2y) =limy→∞(4y2+7y−4y2)(√4y2+7y+2y) =limy→∞(7y)(√4y2−7y+2y) =limy→∞7(√4+7y+2) =72+2=74

${lim}_{x \to - \infty} (\sqrt{4 x^{2} - 7 x} + 2 x)$