# How do you find the limit of (x+4x^2+sinx)/(3x) as x approaches 0?

Mar 12, 2015

If you try to calculate the limit directly you'll get the form $\frac{0}{0}$ that is not good!

Try using de l'Hospital Rule deriving separately numerator and denominator and then do the limit; you get:

${\lim}_{x \to 0} \frac{1 + 8 x + \cos \left(x\right)}{3} = \frac{2}{3}$

Mar 13, 2015

To evaluate this limit without l'Hopital's rule, re-write the expression as:

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

We do this because we can now re=write the limit:

${\lim}_{x \rightarrow 0} \frac{x + 4 {x}^{2} + \sin x}{3 x} = \frac{1}{3} {\lim}_{x \rightarrow 0} \left(\frac{x + 4 {x}^{2}}{x} + \sin \frac{x}{x}\right)$

$= \frac{1}{3} {\lim}_{x \rightarrow 0} \left(\frac{x \left(1 + 4 x\right)}{x} + \sin \frac{x}{x}\right)$

$= \frac{1}{3} {\lim}_{x \rightarrow 0} \left(\left(1 + 4 x\right) + \sin \frac{x}{x}\right)$

$= \frac{1}{3} \left(1 + 1\right) = \frac{2}{3}$