$\lim\limits_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$

Question

Can anyone help me solve this? I know the answer is 4, but I don't really know how do I find the biggest power of $x$ when there's a square root.

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

Answer 1

In a limit tending to $+\infty$ or $-\infty$, the highest-degree term (say $cx^n$) of a polynomial subexpression dominates. Among other things, this means that square roots of such polynomials may be replaced with $\operatorname{sgn}(c)\sqrt{|c|}x^{n/2}$ when evaluating the limit, assuming it is well-defined.

For the given example: $$\lim_{x\to\infty}\frac{\sqrt{4x^2+\color{blue}{x^4}}+3x^2}{x^2-5x}=\lim_{x\to\infty}\frac{\color{blue}{x^2}+3x^2}{x^2-5x}=\lim_{x\to\infty}\frac{\color{green}{4x^2}}{\color{green}{x^2}-5x}=4$$

Answer 2

HINT

Take $x^2$ as a common factor and simplify the quotient.

The limit will be $4$

Answer 3

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

Now as $x\to\infty$ $4/x^2,5/x$ will go to $0$. Thus we have $4$.

Answer 4

As an alternative, note that

$$\frac{\sqrt{x^4}+3x^2}{x^2-5x}=\frac{4x^2}{x^2-5x}\le\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\le\frac{\sqrt{(x^2+2)^2}+3x^2}{x^2-5x}=\frac{4x^2+2}{x^2-5x}$$

thus for squeeze theorem

$$\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\to4$$