Question

How do you find the limit of `(x-4)/(sqrt(x-3)-sqrt(5-x))` as x approaches 4?

Answer 1

The answer is: `1`.

This limit shows up in the `0/0` indecision form, so we have to rationalise.

`lim_(xrarr4)(x-4)/(sqrt(x-3)-sqrt(5-x))=`

`=lim_(xrarr4)[(x-4)/(sqrt(x-3)-sqrt(5-x))*(sqrt(x-3)+sqrt(5-x))/(sqrt(x-3)+sqrt(5-x))]=`

`=lim_(xrarr4)[(x-4)(sqrt(x-3)+sqrt(5-x))]/(x-3-5+x)=`

`=lim_(xrarr4)[(x-4)(sqrt(x-3)+sqrt(5-x))]/(2x-8)=`

`=lim_(xrarr4)[(x-4)(sqrt(x-3)+sqrt(5-x))]/[2(x-4)]=`

`=lim_(xrarr4)[(sqrt(x-3)+sqrt(5-x))]/2=(sqrt(4-3)+sqrt(5-4))/2=1`.

Answer 2

Rationalize the denominator by multiplying the fraction by `1` in the form:

`(sqrt(x-3)+sqrt(5-x))/(sqrt(x-3)+sqrt(5-x))`

This works because `(a-b)(a+b) = a^2-b^2`,

so `(sqrta-sqrtb)(sqrta+sqrtb) = a-b`. It looks like this:

`((x-4))/((sqrt(x-3)-sqrt(5-x))) ((sqrt(x-3)+sqrt(5-x)))/((sqrt(x-3)+sqrt(5-x)))`

`=((x-4) (sqrt(x-3)+sqrt(5-x)))/((x-3)-(5-x)`

`=((x-4) (sqrt(x-3)+sqrt(5-x)))/(2x-8)`

`=((x-4) (sqrt(x-3)+sqrt(5-x)))/(2(x-4))`

`=(sqrt(x-3)+sqrt(5-x))/2`

So, we have:

`lim_(x rarr 4) (x-4)/(sqrt(x-3)-sqrt(5-x))=lim_(xrarr4) (sqrt(x-3)+sqrt(5-x))/2`

`= 2/2=1`