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I'm trying to use this function to compute the derivative. $$\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{(x+h)-x}$$ But I'm stuck when I attempt to find the derivative of $f(2)$ for $f(x) =x^2$ The power rule suggests that I should obtain $nx^{n-1}$ as the result but I have no idea how to compute this limit: $$\lim_{h\rightarrow0}\frac{2^{2+h}-2^2}{(2+h)-2}$$ [updates]

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It should be $(2+h)^2$ not $2^{2+h}$ – Ethan Apr 21 '13 at 10:05
@Ethan Aha! Thank you! But I still want to know how should that limit be computed. – Bob Apr 21 '13 at 10:11
up vote 0 down vote accepted

Hint: $f(2+h) = (2+h)^2 = 4+4h+h^2$ rather than $2^{2+h}$.

Note that $\displaystyle\lim_{h\to 0}\frac{2^{h+2}-2^2}{(2+h)-2} = \left.\frac{\mathrm{d}}{\mathrm{d}x}\right|_{x=2} 2^x = \left.\frac{\mathrm{d}}{\mathrm{d}x}\right|_{x=2} e^{x\ln(2)} = \ln(2)2^x|_{x=2} = 4\ln(2)$. Proving this limit from the definition would probably be very messy.

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Thanks for pointing my careless mistake out! But I still want to know how to compute this limit: $$\lim_{h\rightarrow0}\frac{2^{2+h}-2^2}{(2+h)-2}$$ – Bob Apr 21 '13 at 10:12
So you actually want $g^\prime(2)$ for $g\colon x\mapsto 2^x$? – Abel Apr 21 '13 at 10:13
Nope, I know how solve that already. But now I still want compute this limit: $$\lim_{h\rightarrow0}\frac{2^{2+h}-2^2}{(2+h)-2}$$ – Bob Apr 21 '13 at 10:18
Yes, that is the definition of $\left.\frac{\mathrm{d}}{\mathrm{d}x}\right|_{x=2}(2^x)$. – Abel Apr 21 '13 at 10:20

So what you have is precisely $$\lim_{h \to 0 } 4 \cdot\frac{ 2^{h} -1}{h}$$ and this is nothing but the derivative of $f(x) = 2^{x}$ evaluated at $0$.

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