Question

How do you find `a` and `b` if `\lim _ { x \rightarrow 2} \frac { a \sqrt { x + 2} - b } { x - 2} = 1`?

Answer 1

Please see below.

Explanation:

In order to have `lim_(xrarr2)(asqrt(x+2)+b)/(x-2) = 1`,

we must have initial form indeterminate `0/0`.

So at `x = 2` we have `asqrt(2+2)+b = 0`,

so `2a+b=0` and `b = -2a`

Returning to the limit and substituting for `b`, we get

`lim_(xrarr2)(asqrt(x+2)-2a)/(x-2) = 1`,

Let's factor out the `a`. (Why? because we can and we have to try something. Will it help? Let's find out.)

`lim_(xrarr2)a(sqrt(x+2)-2)/(x-2) = 1`,

Now what should we do? Stop thinking about what we should do and ask yourself what we could do, then try it and see if it helps. If it helps, good. If not try something else.

Often when we see quotients like this, we rationalize the numerator. So, let's try that.

`lim_(xrarr2)a(sqrt(x+2)-2)/(x-2) = lim_(xrarr2)(a(sqrt(x+2)-2)(sqrt(x+2)+2))/((x-2)(sqrt(x+2)+2))`

`= lim_(xrarr2)(a(x+2-4))/((x-2)(sqrt(x+2)+2))`

`= lim_(xrarr2)a/(sqrt(x+2)+2)`

This is not an indeterminate form, so we can evaluate the limit and set it to `1` and finish.

`= a/(sqrt((2)+2)+2) = a/4`

Now, `a/4 = 1` if and only if `a = 4`

and we had `b = -2a`, so we want `b = -8`

And just for fun, here is the computer generated graph of

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

graph{(4sqrt(x+2)-8)/(x-2) [-3.416, 7.68, -2.11, 3.44]}