Limit of (sqrt(16x^4+64x^2)) /(2x^2_4)

[karush]

$$\lim_{{x}\to{\infty}}\frac{\sqrt{16{x}^{4}+64 {x}^{2} }+x^2}{2x^{2} - 4}=\frac{5}{2}$$

I tried to solve this by dividing all terms by$x^4$ but then the denomator will go zero.

 

[Greg]

Divide all terms by $x^2$. Don't forget to square it (so it becomes $x^4$) when you bring it under the radical.

 

[karush]

$$\lim_{{x}\to{\infty}}\frac{\sqrt{16{x}^{4}+64 {x}^{2} }+x^2} {2x^{2} - 4}=\frac{5}{2}$$ Factor out $x^4$ in radical $$\frac{\sqrt{{x}^{4}\left(16-\frac{64}{{x}^{2}}\right)}+{x}^{2}} {2{x}^{2}+4}$$

David by $x^2 $
$$\frac{\frac{{x}^{2}}{{x}^{2}}\sqrt{16-\frac{64}{{x}^{2 }}}
+\frac{{x}^{2}}{{x}^{2}}}{\frac{2{x}^{2}}{{x}^{2}} +\frac{4}{{x}^{2}}}
\implies\frac{\sqrt{16-\frac{64}{{x}^{2}}}+1}{2 +\frac{4}{{x}^{2}}}$$
$$x\to\infty$$
$$\frac{\sqrt{16}+1}{2+0}=\frac{5}{2
}$$