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I've been learning calculus for last 2 weeks since I've got an exam quite soon. I've been doing many exercises on limits, but now I'm stuck in a problem

Can someone please help me out on this problem, I don't know how to approach this problem

$$\lim_{x\to\infty}\frac{\sqrt{9x^2+2x-3}}{(8x^5-6x+1)^{\frac 13}}$$

This is the math problem I dont know how to solve.

Thanks

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  • $\begingroup$ In particular this is a limit at $\infty$, you should update the title. To take an exam after only 2 weeks is not a simple task! $\endgroup$ – gimusi Feb 12 '18 at 14:35
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This is a pretty straightforward calculus problem, and it's standard method is quite popular.

$$\begin{align}\lim_{x\to\infty}\frac{\sqrt{9x^2+2x-3}}{(8x^5-6x+1)^{\frac 13}} \\ &= \lim_{x\to\infty}\frac{\frac 1x\sqrt{9x^2+2x-3}}{\frac 1x(8x^5-6x+1)^{\frac 13}} \\ &= \lim_{x\to\infty}\frac{\sqrt{9+2/x-3/x^2}}{(8x^2-6/x^2+1/x^3)^{\frac 13}} \end{align}$$

the numerator tends to $3$, while the denominator tends to infinity. Hence, the limit is zero.

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$\textbf{HINT:}$ $$ 9x^2+2x-3 \sim_\infty 9x^2 \qquad \mbox{ and } \qquad 8x^5-6x+1 \sim_\infty 8x^5$$

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The standard way to calculate this kind of limits is to collect the leading terms from numerator and denominator as follow

$$\frac{\sqrt{9x^2+2x-3}}{(8x^5-6x+1)^{\frac 13}}=\frac{x}{x^{\frac53}}\cdot\frac{\sqrt{9+2/x-3/x^2}}{(8-6/x^4+1/x^5)^{\frac 13}}\to 0\cdot \frac32=0$$

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