Thread: How would you approach this

1. How would you approach this

Find the values for the constants a and b such that
$\lim[x->o]{\frac{\sqrt{a+bx}-\sqrt{3}}{x}}=\sqrt{3}$

As the title suggests, I'm stuck as to how one would approach this problem.
I can remove the roots, but I don't know what I should do to remove the x from the denominator. I'm thinking that b = all real, a = 12 but I can't prove it.
Thoughts? Suggestions?

2. Originally Posted by Guess992
Find the values for the constants a and b such that
$\lim[x->o]{\frac{\sqrt{a+bx}-\sqrt{3}}{x}}=\sqrt{3}$

As the title suggests, I'm stuck as to how one would approach this problem.
I can remove the roots, but I don't know what I should do to remove the x from the denominator. I'm thinking that b = all real, a = 12 but I can't prove it.
Thoughts? Suggestions?
$a = 3$ , because any other value will make the limit nonexistent.

$\lim_{x \to 0} \frac{\sqrt{3+bx}-\sqrt{3}}{x} \cdot \frac{\sqrt{3+bx} + \sqrt{3}}{\sqrt{3+bx} + \sqrt{3}} =\sqrt{3}$

$\lim_{x \to 0} \frac{(3+bx)-3}{x(\sqrt{3+bx} + \sqrt{3})} = \sqrt{3}$

$\lim_{x \to 0} \frac{bx}{x(\sqrt{3+bx} + \sqrt{3})} = \sqrt{3}$

$\lim_{x \to 0} \frac{b}{\sqrt{3+bx} + \sqrt{3}} = \sqrt{3}$

$\frac{b}{2\sqrt{3}} = \sqrt{3}$

$b = 6$

3. Could you explain how it fails to exist if a =/= 3? I don't quite follow how you get to that point.

4. Originally Posted by Guess992
Could you explain how it fails to exist if a =/= 3? I don't quite follow how you get to that point.
let a = any value except 3 and determine the limit as $x \to 0$ ... see what happens.

5. As x goes to 0, the numerator goes to 0, the numerator goes to $\sqrt{a}- \sqrt{3}$ while the denominator goes to 0. If the numerator does NOT go to 0, then the fraction gets larger and larger without bound. There would be no limit.