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Lim x→∞x(√(x^2+4x)-x-2) equals Single correct option
lim x → ∞ x ( x 2 + 4 x − x − 2 ) \displaystyle\lim_{x\to\infty}x\left(\sqrt{x^2+4x}-x-2\right) x → ∞ lim x ( x 2 + 4 x − x − 2 ) equals
A. 2 2 2 B. − 2 -2 − 2 C. − 4 -4 − 4 D. 0 0 0 Show answer and solution Answer: (B) − 2 -2 − 2
Write
x 2 + 4 x = x 1 + 4 x \sqrt{x^2+4x}=x\sqrt{1+\dfrac4x} x 2 + 4 x = x 1 + x 4 and expand for large
x x x using
1 + u = 1 + u 2 − u 2 8 + ⋯ \sqrt{1+u}=1+\dfrac u2-\dfrac{u^2}8+\cdots 1 + u = 1 + 2 u − 8 u 2 + ⋯ with
u = 4 x u=\dfrac4x u = x 4 :
x 2 + 4 x = x ( 1 + 2 x − 2 x 2 + ⋯ ) = x + 2 − 2 x + ⋯ \sqrt{x^2+4x}=x\left(1+\dfrac2x-\dfrac2{x^2}+\cdots\right)=x+2-\dfrac2x+\cdots x 2 + 4 x = x ( 1 + x 2 − x 2 2 + ⋯ ) = x + 2 − x 2 + ⋯ . Then
x 2 + 4 x − x − 2 = − 2 x + O ( x − 2 ) \sqrt{x^2+4x}-x-2=-\dfrac2x+O\!\left(x^{-2}\right) x 2 + 4 x − x − 2 = − x 2 + O ( x − 2 ) and multiplying by
x x x gives
− 2 + O ( 1 / x ) → − 2 -2+O(1/x)\to-2 − 2 + O ( 1/ x ) → − 2 . Alternatively rationalise:
x 2 + 4 x − ( x + 2 ) = x 2 + 4 x − ( x + 2 ) 2 x 2 + 4 x + x + 2 = − 4 x 2 + 4 x + x + 2 \sqrt{x^2+4x}-(x+2)=\dfrac{x^2+4x-(x+2)^2}{\sqrt{x^2+4x}+x+2}=\dfrac{-4}{\sqrt{x^2+4x}+x+2} x 2 + 4 x − ( x + 2 ) = x 2 + 4 x + x + 2 x 2 + 4 x − ( x + 2 ) 2 = x 2 + 4 x + x + 2 − 4 , and
x x x times this tends to
− 4 2 = − 2 \dfrac{-4}{2}=-2 2 − 4 = − 2 .
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