Question

What is the value of `lim_(xrarr0)((x+32)^(1/5)-2)/x`?

Answer 1

`1/80`

Explanation:

The first step is to just try and plug in `0` to see what we get. I might just quickly note that `root(5)(32)=32^(1/5)=2`.
`((0+32)^(1/5)-2)/0=(2-2)/0=0/0`

This result doesn't help us in computing the limit. There is however something we can do to evaluate the limit. We can use a nice technique called L'Hôpital's rule. It says:
If `lim_(x->k)f(x)=0` and `lim_(x->k)g(x)=0` and `lim_(x->k)(f'(x))/(g'(x))=P`
Then `lim_(x->k)f(x)/g(x)=P`

What this means is that if we have a limit with the indeterminate form `0/0`, we can take the derivative of both the numerator and the denominator and we'll still end up with the same answer.

We can compute the derivatives using the chain rule and power rule:
`d/dx((x+32)^(1/5)-2)=1/5(x+32)^(-4/5)`

`dx/dx=1`

So our limit becomes:
`lim_(x->0)(1/5(x+32)^(-4/5))/1=lim_(x->0)1/(5(x+32)^(4/5))`

Now we can plug in `0` to get the answer:
`1/(5(0+32)^(4/5))=1/(5*32^(4/5))=1/(5*16)=1/80`