What is the limit of  ((1 + 3 x) / (2 + 4 x^2 + 2 x^4))^3  as x approaches 1?

Oct 5, 2016

$y = \frac{1}{8}$

Explanation:

To make the computes easier we can rewrite the expression like

$f \left(x\right) = {\left(\frac{1 + 3 x}{2 + 4 {x}^{2} + 2 {x}^{4}}\right)}^{3} = {\left(\frac{1 + 3 x}{2 \cdot \left(1 + 2 {x}^{2} + {x}^{4}\right)}\right)}^{3} = \frac{1}{2} ^ 3 {\left(\frac{1 + 3 x}{{x}^{4} + 2 {x}^{2} + 1}\right)}^{3} = \frac{1}{8} {\left(\frac{1 + 3 x}{{\left({x}^{2} + 1\right)}^{2}}\right)}^{3}$

You have to find

${\lim}_{x \to 1} f \left(x\right) = {\lim}_{x \to 1} \frac{1}{8} {\left(\frac{1 + 3 x}{{\left({x}^{2} + 1\right)}^{2}}\right)}^{3} = \frac{1}{8} {\lim}_{x \to 1} {\left(\frac{1 + 3 x}{{\left({x}^{2} + 1\right)}^{2}}\right)}^{3}$

Since $f \left(x\right)$ is continuos in $x = 1$, because ${\left({x}^{2} + 1\right)}^{2} \ne 0 \forall x$

to evaluate the limit you could simply substitute $x = 1$

${\lim}_{x \to 1} f \left(x\right) = \frac{1}{8} {\left(\frac{1 + 3 \left(1\right)}{{\left({\left(1\right)}^{2} + 1\right)}^{2}}\right)}^{3} = \frac{1}{8} {\left(\frac{1 + 3}{1 + 1} ^ 2\right)}^{3} =$
$= \frac{1}{8} {\left(\frac{\cancel{4}}{\cancel{4}}\right)}^{3} = \frac{1}{8} \cdot 1 = \frac{1}{8}$

graph{((1+3x)/(2+4x^2+2x^4))^3 [0.9017, 1.1578, 0.0647, 0.1928]}