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!
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 $\displaystyle \lim_{h \to 0} \frac {3^h - 1}h$ or equivalently, $\displaystyle \lim_{h \to 0} \frac {e^{h \ln 3} - 1}h$. (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 $\displaystyle e^x$) and not by evaluating the limit directly.
we use the fact that $\displaystyle e^{\ln X} = X$ and $\displaystyle \log_a x^n = n \log_a x$
so, $\displaystyle \frac d{dx}a^x = \frac d{dx} e^{x \ln a} = \bigg( \frac d{dx} x \ln a \bigg) \cdot e^{x \ln a}$ by the chain rule. which simplifies to $\displaystyle \ln a \cdot a^x = a^x \ln a$ for $\displaystyle a > 0$
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 $\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x - h)}{2h}$. this is equivalent to the previous definition. we just used the points $\displaystyle (x + h)$ and $\displaystyle (x - h)$ as opposed to $\displaystyle x$ and $\displaystyle (x + h)$.
so, $\displaystyle \lim_{h \to 0} \frac {3^{4 + h} - 3^4}h = \lim_{h \to 0} \frac {3^{4 + h} - 3^{4 - h}}{2h}$
$\displaystyle = 3^4 \cdot \lim_{h \to 0} \frac {3^h - 3^{-h}}{2h}$
$\displaystyle = 3^4 \cdot \lim_{h \to 0} \frac {e^{h \ln 3} - e^{- h \ln 3}}{2h}$
$\displaystyle = 3^4 \cdot \lim_{h \to 0} \frac {\sinh (h \ln 3)}h$
$\displaystyle = 3^4 \cdot \ln 3 \cdot \lim_{h \to 0} \frac {\sinh (h \ln 3)}{h \ln 3}$
$\displaystyle = 3^4 \cdot \ln 3 \cdot 1$
$\displaystyle = 3^4 \cdot \ln 3$