evaluating limits

**sarabolha** · September 13th 2008, 03:22 PM

evaluate:

lim 3^(4+h) - 3^(4)
h->0 h

i think i remember something from calc that had to do with the definition of a derivative, and a variation of the formula looking something similiar to this... any help?? i dont know where to begin...

thanks!

**Plato** · September 13th 2008, 03:32 PM

Take note of $\frac{{3^{h + 4} - 3^4 }}{h} = 3^4 \frac{{3^h - 1}}{h}$ .
Now the problem reduces to finding if $f(x)=3^x$ then $f'(0)=?$

**sarabolha** · September 13th 2008, 03:44 PM

okay, that does help a ton, and i know the answer is 3^4 * ln(3).

but i know that from memorization, would you mind explaining how that is derived?

**Plato** · September 13th 2008, 03:59 PM

$f(x) = 3^x \, \Rightarrow \,f'(0) = \lim _{h \to 0} \frac{{f(h) - f(0)}}<br /> {h}$

**sarabolha** · September 13th 2008, 04:14 PM

Originally Posted by Plato

$f(x) = 3^x \, \Rightarrow \,f'(0) = \lim _{h \to 0} \frac{{f(h) - f(0)}}<br /> {h}$

how did i come up with ln3? i'm sorry but can you explain it in simpler terms i'm not as familiar with math as i used to be

**Chop Suey** · September 13th 2008, 05:04 PM

$\frac{d}{dx} a^x = a^x \ln{a}$

**Jhevon** · September 13th 2008, 06:37 PM

Originally Posted by sarabolha

evaluate:

lim 3^(4+h) - 3^(4)
h->0 h

i think i remember something from calc that had to do with the definition of a derivative, and a variation of the formula looking something similiar to this... any help?? i dont know where to begin...

thanks!

finding the limit directly is a pain. unless you know how to find $\lim_{h \to 0} \frac {3^h - 1}h$ or equivalently, $\lim_{h \to 0} \frac {e^{h \ln 3} - 1}h$ . (TPH or Kriz can probably help you with that

)

the rule is genrally derived by using the chain rule (once we know the derivative of $e^x$ ) and not by evaluating the limit directly.

we use the fact that $e^{\ln X} = X$ and $\log_a x^n = n \log_a x$

so, $\frac d{dx}a^x = \frac d{dx} e^{x \ln a} = \bigg( \frac d{dx} x \ln a \bigg) \cdot e^{x \ln a}$ by the chain rule. which simplifies to $\ln a \cdot a^x = a^x \ln a$ for $a > 0$

so recognizing that it is the derivative limit and running the above proof is the way to go

an alternative to find the limit without this, is to use a slightly modified version of the difference quotient, and some knowledge of hyperbolic functions.

let $f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x - h)}{2h}$ . this is equivalent to the previous definition. we just used the points $(x + h)$ and $(x - h)$ as opposed to $x$ and $(x + h)$ .

so, $\lim_{h \to 0} \frac {3^{4 + h} - 3^4}h = \lim_{h \to 0} \frac {3^{4 + h} - 3^{4 - h}}{2h}$

$= 3^4 \cdot \lim_{h \to 0} \frac {3^h - 3^{-h}}{2h}$

$= 3^4 \cdot \lim_{h \to 0} \frac {e^{h \ln 3} - e^{- h \ln 3}}{2h}$

$= 3^4 \cdot \lim_{h \to 0} \frac {\sinh (h \ln 3)}h$

$= 3^4 \cdot \ln 3 \cdot \lim_{h \to 0} \frac {\sinh (h \ln 3)}{h \ln 3}$

$= 3^4 \cdot \ln 3 \cdot 1$

$= 3^4 \cdot \ln 3$

Thread: evaluating limits

LinkBack

Thread Tools

Display

evaluating limits

Similar Math Help Forum Discussions

Evaluating Limits

Evaluating Limits

Evaluating limits

evaluating limits

Help evaluating limits

Search Tags