Question

Find the limits of $\frac{\log (1+x^2+x^4)}{3x^2(1-2x)},$ when $x=0$.

Answer 1

To find the limits of $\frac{\log (1+x^2+x^4)}{3x^2(1-2x)}$, when $x=0$

i.e. $\underset{x\to 0}{lim}\frac{\log (1+x^2+x^4)}{3x^2(1-2x)}$

$=\underset{x\to 0}{lim}\frac{(x^2+x^4)-\frac{1}{2}(x^2+x^4{)}^2+......}{3x^2(1-2x)}$ ...... [Using the Logarithmic series expansion]

$=\underset{x\to 0}{lim}\frac{(x^2+x^4)}{3x^2(1-2x)}$ .... [Neglecting higher terms]

$=\underset{x\to 0}{lim}\frac{(1+x^2)}{3(1-2x)}$

$=\frac{1}{3}$