I completely forgot how to do this...?

Question

I haven't done limits in a while and I was looking at some "fun problems"...out of my textbook...

lim ( cubed root of (1+mx) -1)/x

x->0

..where m is constant...

the cubed root is only on the 1+mx...

Answer 1

The classical solution is based on the De' l 'hospital rule, and take the derivatives of the numerator and denominator

In

lim ((1+mx)^(1/3) -1)/ x

x->0

it is

lim((1+mx)^1/3 -1) = 0

x->0

and

lim(x) = 0

Thus,

lim ((1+mx)^(1/3) -1)/ x = lim ((1+mx)^1/3 - 1) ' / (x)'

x->0                              x->0

lim(-2*m / (3 (1+mx)^2/3) = -sign(m) * Infinity

x->0

Answer 2

Use L'hopital's rule because when you directly substitute in 0 for x, you get 0/0 which is valid to use the rule.

Therefore, take the derivative of the top and the bottom with respect to x.

Take derivative of numerator and denominator:

Numerator

(1+mx)^(1/3)-1 dx

(m/3)(1+mx)^(-2/3)

m/(3(1+mx)^(2/3))

Denominator

x dx

1

So, the new function is m/(3(1+mx)^(2/3)

Now take the limit of the new function as x approaches 0.

Plug in 0.

m/(3(1)^(2/3))

m/3    <------- Final Answer