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$$\lim_{x\to16}\frac{\sqrt[4]x-2}{x-16}$$

My suspicion is that I need to find the conjugate to get this in a more factorable form. From there, I think I can plug in $16$ to find the limit. The only issue is that I'm not quite sure how to factor this. This is my first Calc class and I've never factored anything quite like this before. I found the following format for factoring this sort of problem:

$$(a-b)\left(a^3+a^2b+ab^2+b^3\right) = a^4-b^4$$

But I'm pretty confused in terms of how to apply it (do I apply it to the numerator and then try to get the denominator into a similar form to cancel stuff out?). Does $a = x^\frac14$, and $b = 2$? I feel like the toughest part of this problem is the factoring, not the Calculus.

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Hint: Use the identity $a^2-b^2=(a-b)(a+b)$ to factor the difference of two squares.

You can factor the denominator by recognizing that $x-16$ is the difference of two squares: $x-16=(\sqrt x-4)(\sqrt x+4)$. Then factor $\sqrt x - 4$ by recognizing it as the difference of two squares...

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  • $\begingroup$ This worked for me - I was able to factor the denominator and then cancel from the top, leaving me with 1/32 after I plugged in 16. Thank you! $\endgroup$ – Chris T Aug 31 '16 at 0:27
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The limit is an indeterminate form. If you evaluate the limit you get $\frac{0}{0}$. So you can apply L'Hopital's rule.

$$ \lim_{x\rightarrow16}\frac{\sqrt[4]{x}-2}{x-16}=\lim_{x\rightarrow16}\frac{\frac{d}{dx}\left(\sqrt[4]{x}-2\right)}{\frac{d}{dx}\left(x-16\right)}. $$

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The standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ comes to the rescue. Here a = 16, n = 1/4 and hence the desired limit is $\dfrac{1}{4}\cdot 16^{-3/4} = \dfrac{1}{32}$

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