What is the best way to solve problem like these ↓ ? (Limits to infinity)

Question

Problems such as

When $x$ $x$ approaches positive infinity, find the limit of

$\frac{3 x - 1}{\sqrt{x^{2} - 6}}$ $(3x-1)/(sqrt(x^2-6$

$\frac{\sqrt{4 x^{2} + 4 x}}{4 x + 1}$ $(sqrt(4x^2+4x))/(4x+1)$

I know that you can solve by substituting values of $x$ $x$ as it gets closer to infinity, but is there a way to solve, like factorising, formula, or a general rule?

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Answer 1 · 2017-06-22T06:41:52

As a general rule convert a polynomial $f (x)$ $f(x)$ to $g (\frac{1}{x})$ $g(1/x)$
(A) $3$ $3$ and (B) $\frac{1}{2}$ $1/2$

$L t_{x \to \infty} \frac{3 x - 1}{\sqrt{x^{2} - 6}}$ $Lt_(x->oo)(3x-1)/sqrt(x^2-6)$

$L t_{x \to \infty} \frac{3 - \frac{1}{x}}{\sqrt{1 - \frac{6}{x^{2}}}}$ $Lt_(x->oo)(3-1/x)/sqrt(1-6/x^2)$

= $\frac{3}{\sqrt{1}} = 3$ $3/sqrt1=3$
graph{(3x-1)/sqrt(x^2-6) [2.07, 19.55, -0.37, 8.37]}

$L t_{x \to \infty} \frac{\sqrt{4 x^{2} + 4 x}}{4 x + 1}$ $Lt_(x->oo)sqrt(4x^2+4x)/(4x+1)$

$L t_{x \to \infty} \frac{\sqrt{1 + \frac{2}{x}}}{2 + \frac{1}{2} x}$ $Lt_(x->oo)sqrt(1+2/x)/(2+1/2x)$

= $\frac{1}{2}$ $1/2$
graph{sqrt(4x^2+4x)/(4x+1) [-0.152, 4.218, -0.51, 1.675]}

As a general rule convert a polynomial $f (x)$ $f(x)$ to $g (\frac{1}{x})$ $g(1/x)$