Question

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

Answer 1

If you try to calculate the limit directly you'll get the form `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->0)(1+8x+cos(x))/3=2/3`

Answer 2

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

`(x+4x^2+sinx)/(3x)=1/3((x+4x^2)/x+sinx/x)`

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

`lim_(xrarr0)(x+4x^2+sinx)/(3x)=1/3lim_(xrarr0)((x+4x^2)/x+sinx/x)`

`=1/3lim_(xrarr0)((x(1+4x))/x+sinx/x)`

`=1/3lim_(xrarr0)((1+4x)+sinx/x)`

`=1/3(1+1)=2/3`