Help with limit of radical function

Question

$$\lim_{x \to \infty} \frac{\sqrt{4x^{4}+3}}{5x^2+3}$$

$$= \lim_{x \to \infty} \frac{(4x^{4}+3)^{1/2}}{5x^2+3}$$

$$= \lim_{x \to \infty} \frac{(\frac{4x^{4}}{x^{1/2}} +\frac{3}{x^{1/2}})^{1/2} }{5+\frac{3}{x^2}}$$

$$= \lim_{x \to \infty} \frac{2x+\frac{\sqrt{3}}{x}}{5+\frac{3}{x^2}}$$

The answer is $\frac{2}{5}$. Why is this the answer? I guess what confuses me is this part: $(4x^2)^{1/2}= 2x^1=2x.$ Is that not so? How are they getting $2x^0=2$ from that?

Answer 1

When you divided by $x^2$ in the numerator and denominator and moved it inside the radical, you should have been dividing by $x^4$, not $x^{1/2}$. Consequently, you should have $$\lim_{x\to\infty} \frac{\sqrt{4+3x^{-4}}}{5+3x^{-2}}$$ Which does, indeed, evaluate to $2/5$

Answer 2

For $x\rightarrow\infty$

$$ \frac{\sqrt{4x^4+3}}{5x^2+3}\sim \frac{\sqrt{4x^4}}{5x^2}=\frac{2x^2}{5x^2}=\frac{2}{5} $$