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I can't evaluate the limit of the following expression: $$ \lim_{x \to \infty} \frac{\sqrt{2x + 3}}{3x - 1} $$ Taking the square root in the numerator is confusing me.

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  • $\begingroup$ Divide top and bottom by $x$... $\endgroup$ – abiessu Nov 16 '13 at 18:32
  • $\begingroup$ Note that there is no equation. You are trying to find the limit of the expression. $\endgroup$ – Sammy Black Nov 16 '13 at 18:34
  • $\begingroup$ Presumably you know what the thing is like when $x$ is huge. Like $x\approx 10^{12}$. The top is $\lt 2\times 10^6$, the bottom is $\approx 3\times 10^{12}$, the ratio is real close to $0$. $\endgroup$ – André Nicolas Nov 16 '13 at 20:01
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Divide the numerator and denominator by $x$, and note that $x = \sqrt {x^2},\;$ since ($x\to +\infty \implies x>0$).

$$\lim_{x\to \infty} \dfrac{\sqrt {2x+ 3}}{3x - 1} = \lim_{x\to \infty} \dfrac{\frac{\sqrt {2x+ 3}}{\sqrt {x^2}}}{\frac{3x - 1}{x}} =\lim_{x \to \infty} \dfrac{\sqrt {\frac{2}{x} + \frac 3{x^2}}}{3 - \frac 1x} = \dfrac {0}3 = 0$$

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First note that for $x > 3$ we have $2x+3 < 3x$ and $3x-1 > 2x$. Hence, we have for $x > 3$, $$0 < \dfrac{(2x+3)^{0.5}}{3x-1} < \dfrac{(3x)^{0.5}}{2x} = \dfrac{\sqrt3}2 \dfrac1x$$ Now conclude what you want.

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