- Thread starter JamesM0025
- Start date

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 \(\displaystyle (h^2 + 15) - (h + 15) = h^2 - h\).@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.

-Dan

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 \(\displaystyle \sqrt{(h^2+15)(h+15)}\)

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

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

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

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

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

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

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

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

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

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

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

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