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I tried with Wolfram Alpha, but it fails to provide a step-by-step solution and gives me only the answer.

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  • $\begingroup$ Taylor series if you know them, rationalization (multiply by $\frac{\sqrt{9x^2+5}+3x}{\sqrt{9x^2+5}+3x}$) if you don't? $\endgroup$ – Clement C. Nov 21 '16 at 15:09
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Rewrite $\sqrt{9x^2+5}-3x$ as a fraction.

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Alternative hint: set $x=1/\sqrt{t}$ and compute $$ \lim_{t\to0^+}\frac{1}{\sqrt{t}}\left(\sqrt{\frac{9}{t}+5}-\frac{3}{\sqrt{t}}\right) = \lim_{t\to0^+}\frac{\sqrt{9+5t}-3}{t} $$ which is a simple derivative.

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$\lim_{x\to \infty }\left(x\left(\sqrt{9x^2+5}-3x\right)\right)=\lim_{x\to\infty}\left(x\left(\sqrt{9x^2+5}-3x\right)\frac{\sqrt{9x^2+5}+3x}{\sqrt{9x^2+5}+3x}\right)=\lim_{x\to\infty}\frac{5x}{\sqrt{9x^2+5}+3x}=5/6$

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Try rationalising the factor sqrt(9x^2+5)-3x by sqrt(9x^2+5)+3x. After that simplify a little and make sure that the variable x is in power of 1/x after cancelling out common x from numerator and denominator. You will get the limit to be 5/6.

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