# Stuck on a simplification sqrt h^2+15 * sqrt h+15

#### JamesM0025

I'm not sure what the result of sqrt (h^2+15) * sqrt (h+15) turns into

#### topsquark

Math Team
For any a, b real we have that $$\displaystyle (a + b)(a - b) = a^2 - b^2$$.

In your case you have $$\displaystyle a = \sqrt{h^2 + 15}$$ and $$\displaystyle b = \sqrt{h + 15}$$.

So what is $$\displaystyle a^2 - b^2$$?

-Dan

idontknow

#### romsek

Math Team
Are you sure you're not just supposed to use L'Hopital's rule?

$\lim \limits_{h\to 1} \dfrac{\dfrac{h}{\sqrt{h^2+15}} - \dfrac{1}{2\sqrt{h+15}}}{\dfrac{1}{2\sqrt{h+3}}} = \\~\\ \dfrac{\dfrac 1 4 - \dfrac 1 8}{\dfrac 1 4} = \dfrac 4 8 = \dfrac 1 2$

topsquark

#### JamesM0025

@topsquark Thanks for the help, but I found myself getting lost in the expanded form of it. Which just added to my confusion when I ended up with 128, but the answer is 1/2 lol. If you scroll up, you can see the original question if you have the time.

#### JamesM0025

@romsek It's a review for a midterm on Saturday, but I know in our latest lab when solving limits it said to not use that rule, so it might also be a thing on the midterm .. or maybe it won't be. I've never seen that rule before. I'll have to watch a video.

#### topsquark

Math Team
@topsquark thanks for the help but i found myself getting lost in the expanded form of it. Which just added to my confusion when i ended up with 128 but the answer is 1/2 lol. If you scroll up you can see the original question if you have the time.
I don't understand. I looked at the original and you had left it in the form (a + b)(a - b). The numerator comes out to be $$\displaystyle (h^2 + 15) - (h + 15) = h^2 - h$$.

-Dan

#### bearnutella

@topsquark Thanks for the help, but I found myself getting lost in the expanded form of it. Which just added to my confusion when I ended up with 128, but the answer is 1/2 lol. If you scroll up, you can see the original question if you have the time.
When multiplying binomials with the same root, you can just multiply the two normally and copy the degree of the root =))
so $$\displaystyle \sqrt{h^2+15}*\sqrt{h+15}$$ will become $$\displaystyle \sqrt{(h^2+15)(h+15)}$$

#### mathman

Forum Staff
Let $x=h-1$, then the problem is simpler $\lim_{x\to 0}\frac{\sqrt{16+2x+x^2}-\sqrt{16+x}}{\sqrt{4+x}-2}$. Expanding the terms to first order in $x$ leads to $\frac{1}{2}$ as the limit. (Numerator $\approx \frac{x}{2}$ and denominator $\approx x$.)

#### skeeter

Math Team
$\dfrac{\sqrt{h^2+15} - \sqrt{h+15}}{\sqrt{h+3}-2} \cdot \dfrac{\sqrt{h^2+15} + \sqrt{h+15}}{\sqrt{h^2+15} + \sqrt{h+15}}$

$\dfrac{(h^2+15)-(h+15)}{(\sqrt{h+3}-2)(\sqrt{h^2+15} + \sqrt{h+15})}$

$\dfrac{h(h-1)}{(\sqrt{h+3}-2)(\sqrt{h^2+15} + \sqrt{h+15})} \cdot \dfrac{\sqrt{h+3}+2}{\sqrt{h+3}+2}$

$\dfrac{h(h-1)(\sqrt{h+3}+2)}{[(h+3)-4] (\sqrt{h^2+15}+\sqrt{h+15})}$

$\dfrac{h(\cancel{h-1})(\sqrt{h+3}+2)}{(\cancel{h-1}) (\sqrt{h^2+15}+\sqrt{h+15})}$

$\displaystyle \lim_{h \to 1} \dfrac{h(\sqrt{h+3}+2)}{\sqrt{h^2+15} + \sqrt{h+15}} = \dfrac{1}{2}$

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#### skipjack

Forum Staff
(Numerator $\approx \frac{x}{2}$ and denominator $\approx x$.)
Numerator $\approx \frac{x}{8}$ and denominator $\approx \frac{x}{4}$.