# How do you find the limit of (2x^3 + 6x^2 +5) / (3 + x^3) as x approaches infinity? 2) lim (2x) / (√(x^2 +1)) x--> +∞ 3) lim (2x) / (√(x^2 +1)) x--> -∞?

##### 1 Answer
Aug 1, 2015

${\lim}_{x \to \infty} \frac{2 {x}^{3} + 6 {x}^{2} + 5}{3 + {x}^{3}} = 2$

#### Explanation:

Right from the get-go and without doing any calculations, you can say that the first limit will be equal to $2$.

That happens because you're dealing with the ratio of two polynomials of the same degree, so the limit when $x \to \infty$ will be determined by the ratio of the coefficients that belong to the terms with the highest degree.

In your case, the limit will be determined by the coefficients of the ${x}^{3}$-terms.

${\lim}_{x \to \infty} \frac{\textcolor{red}{2} {x}^{\textcolor{b l u e}{3}} + 6 {x}^{2} + 5}{{x}^{\textcolor{b l u e}{3}} + 3} = \frac{\textcolor{red}{2}}{1} = 2$

Alternatively, you could actually calculate this limit by doing some algebraic manipulation to the numerator and denominator.

More specifically, you could write

$2 {x}^{3} + 6 {x}^{2} + 5 = {x}^{3} \left(2 + \frac{6}{x} + \frac{5}{x} ^ 3\right)$

and

${x}^{3} + 3 = {x}^{3} \left(1 + \frac{3}{x} ^ 3\right)$

Your limit will now take the form

${\lim}_{x \to \infty} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{3}}}} \left(2 + \frac{6}{x} + \frac{5}{x} ^ 3\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{3}}}} \left(1 + \frac{3}{x} ^ 3\right)} = {\lim}_{x \to \infty} \frac{2 + \frac{6}{x} + \frac{5}{x} ^ 3}{1 + \frac{3}{x} ^ 3} = \frac{2 + 0 + 0}{1 + 0} = \textcolor{g r e e n}{2}$

BONUS

You can use the same technique to determine the other two limits

${\lim}_{x \to \infty} \frac{2 x}{\sqrt{{x}^{2} + 1}}$ and ${\lim}_{x \to - \infty} \frac{2 x}{\sqrt{{x}^{2} + 1}}$

However, you must take into account the fact that $\sqrt{{x}^{2}} = | x |$. Here's what I mean by that.

For your first expression, you can write

$\frac{2 x}{\sqrt{{x}^{2} \left(1 + \frac{1}{x} ^ 2\right)}} = \frac{2 x}{\sqrt{{x}^{2}} \cdot \sqrt{1 + \frac{1}{x} ^ 2}} = \frac{2 x}{| x | \cdot \sqrt{1 + \frac{1}{x} ^ 2}}$

This is where the two limits will differ. For $x \to \infty$, you will use the positive values of $x$, so $| x | = x$. On the other hand, when $x \to - \infty$, you will use the negative values of $x$, so $| x | = - x$.

The two limits will thus be

${\lim}_{x \to \infty} \frac{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \cdot \sqrt{1 + \frac{1}{x} ^ 2}} = {\lim}_{x \to \infty} \frac{2}{\sqrt{1 + \frac{1}{x} ^ 2}} = \frac{2}{1} = \textcolor{g r e e n}{2}$

and

${\lim}_{x \to - \infty} \frac{2 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- x}}} \cdot \sqrt{1 + \frac{1}{x} ^ 2}} = {\lim}_{x \to - \infty} \left(- \frac{2}{\sqrt{1 + \frac{1}{x} ^ 2}}\right) = - \frac{2}{1} = \textcolor{g r e e n}{- 2}$