Question I'm not sure I solved right (Calc 2, L'Hopital's rule I believe)

Question

If I could get a thorough explanation of each step or justification, that would be super appreciated!

Find numbers a and b such that:

lim sqrt(ax+b)-2

x→0 ————— = 1

x

I think this is L'Hopital's rule, but I could only solve for a in terms of b and am not sure about my method of solving b for a specific value.

Answer 1

To me this is not necessarily a L'Hospital's rule case since L'Hospital's Ruke applies to only zero/zero or infinity/infinity cases and that isn't true for all values of "a" and "b".

Instead, I would multiply numerator and denominator by the conjugate "√(ax + b) + 2" which would yield:

or

Since you want to be able to resolve the limit (not DNE) then you must be able to cancel the "x" in the denominator with an "x" in the numerator. The only way that happens is if b = 4. So allowing b = 4 and cancelling the "x" from numerator and denominator, we get:

lim as x approaches zero of a/(√(ax + 4) + 2)

Plugging in x = 0 and setting the result equal to 1 allows you to solve for "a" getting a = 4

So a = 4 and b = 4

Turns out the b = 4 does enable L'Hospital's Rule to be used but you don't know that going in.

Answer 2

In order to apply L'Hopital, must show indeterminate form.

Substituting 0 for x means the numerator only equals zero when √b - 2 = 0 => b = 4.

Now applying L'Hopital:

Derivative of numerator:

(1/2) (ax+4)^-1/2 (a) = a/2√(ax+4)

The derivative of the denominator is 1.

Substituting 0 for x in the new numerator:

a/2√4 should equal 1.

a = 4

This Desmos graph sort of confirms:

desmos.com/calculator/liq1agyy9v