Find the limit... ?

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Find the limit... ?

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Steve M

Jan 13, 2017

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Answer 1 · 2017-01-13T01:21:33

$\frac{1}{2}$ $1/2$

Explanation:

$\sqrt{9 x^{2} + 3 x} - 3 x = \frac{9 x^{2} + 3 x - 9 x^{2}}{\sqrt{9 x^{2} + 3 x} + 3 x} = \frac{3 x}{\sqrt{9 x^{2} + 3 x} + 3 x}$ $sqrt(9x^2+3x)-3x=(9x^2+3x-9x^2)/(sqrt(9x^2+3x)+3x)= (3x)/(sqrt(9x^2+3x)+3x)$ but

$\frac{3 x}{\sqrt{9 x^{2} + 3 x} + 3 x} = \frac{3 x}{3 x (\sqrt{1 + \frac{1}{3 x}} + 1)} = \frac{1}{\sqrt{1 + \frac{1}{3 x}} + 1}$ $(3x)/(sqrt(9x^2+3x)+3x)=(3x)/(3x(sqrt(1+1/(3x))+1))=1/(sqrt(1+1/(3x))+1)$

finally

$lim_(x->oo)sqrt(9x^2+3x)-3x=lim_(x->oo)1/(sqrt(1+1/(3x))+1)=1/2$

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