Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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]

share|improve this question
1  
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
add comment

2 Answers

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.

share|improve this answer
    
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
add comment

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 $0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.