Question

How do you find the limit of `(sqrt (1+9x)- sqrt (1-8x))/ x` as x approaches 0?

Answer 1

I found: `17/2`

Explanation:

If you try directly you get `0/0`:

Answer 2

"Rationalize" the numerator.

Explanation:

This is a trick (technique, method) that is very useful in calculus.

If we multiply `(sqrta-sqrtb)(sqrta+sqrtb)`, we get rid of the square roots and can try to simplify.

`(sqrta-sqrtb)(sqrta+sqrtb) = a-b`

In this case:

`(sqrt (1+9x)- sqrt (1-8x))/ x`

We will multiply by `1` in the form `(sqrt (1+9x) + sqrt (1-8x))/ (sqrt (1+9x) + sqrt (1-8x))`

We get:

`((sqrt (1+9x)- sqrt (1-8x)))/ x *((sqrt (1+9x) + sqrt (1-8x)))/ ((sqrt (1+9x) + sqrt (1-8x)))`

`= ((1+9x)-(1-8x))/(x(sqrt (1+9x) + sqrt (1-8x)))`

`= (17x)/(x(sqrt (1+9x) + sqrt (1-8x)))`

`= 17/(sqrt (1+9x) + sqrt (1-8x))`

Now we can evaluate the limit by substitution.

`lim_(xrarr0)(sqrt (1+9x)- sqrt (1-8x))/ x = lim_(xrarr0)17/(sqrt (1+9x) + sqrt (1-8x))`

`= 17/(sqrt(1+9(0))+sqrt(1-8(0)))`

`= 17/(sqrt1+sqrt1) = 17/2`

In a way, we have 'traded' a subtraction in the numerator that goes to `0` for an addition in the denominator that does not go to `0`.