The following excerpt is from Stewart Calculus: Early Transcendentals, 7th Edition.
My question is how does the author arrive at:
$$\lim_{h \to 0}\frac{a^h-1}{h}=f'(0)$$
Understanding the proof of derivative of exponential function.
calculusexponential functionproof-explanation
Related Solutions
The difference is entirely a pedagogical one. Both approaches ultimately cover the same calculus.
“Late transcendentals” is the traditional approach to teaching calculus where the treatment of logarithmic and exponential functions is postponed until after integration is introduced. In the traditional method, the natural logarithm is defined by the formula $$ \ln x \;\underset{\scriptscriptstyle\mathrm{def}}{=}\; \int_1^x \frac{1}{t}\,dt, $$ and $e^x$ is then defined to be the inverse function of $\ln x$. Important properties of $\ln x$, such as that $\ln(xy) = \ln x + \ln y$, are proven from the integral definition using $u$-substitution, and the properties of the exponential function follow from these. Arbitrary powers $a^b$ where $a>0$ and $b\in\mathbb{R}$ are then defined by the formula $$ a^b = e^{b \ln a} $$ and it is shown that this agrees with the existing definition when $b$ is rational.
In the “early transcendentals” method, the logarithmic and exponential functions are introduced shortly after the definition of the derivative. Exponentiation where the power is an arbitrary real number is defined by the formula $$ a^b \;=\; \lim_{q\to b}\, a^q $$ where the limit is taken over the rational numbers. (There are some technical issues in proving that this limit exists, which are often ignored.) An argument is then given that there exists a number $e$ for which $$ \frac{d}{dx}\bigl[e^x\bigr]\biggr|_{x=0} \;=\; 1, $$ though again the presentation of this argument is sometimes less than completely rigorous. It follows that the derivative of $e^x$ is $e^x$, and the natural logarithm is defined as the inverse of the exponential function.
The traditional method has the advantage of being cleaner, and the proofs are simple enough that they can be presented to starting calculus students in a mathematically rigorous way. However, it has several distinct disadvantages:
It is not very intuitive, since the natural logarithm and exponential function are essentially summoned out of thin air by what looks like magic. The “early transcendentals” approach, on the other hand, corresponds much more closely with how we actually think of exponentials and logarithms.
Students who take only a single semester of calculus, which includes most biology majors at some colleges in the United States, do not see the exponential function and natural logarithm. This is a serious problem, because these are among the most important functions for applications in biology.
Even students who take two semesters of calculus learn about exponential and logarithmic functions fairly late, which means they don't have time to get used to these functions.
These arguments are generally considered persuasive enough that most universities in the United States have now adopted calculus books that use the “early transcendentals” approach.
Of course, not everyone agrees with this change, and further arguments can be presented on both sides of the debate. Books using the traditional approach continue to be used at some universities and in some calculus classes. In particular, some universities offer an honors calculus sequence designed specifically for math majors and similar students, and there's a relatively strong argument for using the traditional approach in such a course. My impression is also that the “late transcendentals” approach remains quite popular outside the United States, though my evidence for this is purely anecdotal.
Because the function has changed. Let's do an example:
$$\int_{-1}^1 x\,dx =0$$
because the integrand is odd and the interval is symmetric (you can also check directly).
Let's do a simple $u=x+1$ so that $du=dx$ then the right way we have:
$$\int_0^2 u-1\,du = {1\over 2}(u-1)^2\bigg|_0^2={1\over 2}-{1\over 2}=0$$
but if we fail to change the limits:
$$\int_{-1}^1(u-1)\,du = {1\over 2}(u-1)^2\bigg|_{-1}^1=-2$$
The underlying reason is that integration comes from Riemann sums, the function values depend on the interval of integration. When you change the interval, the heights of the rectangles you use in the definition change (remember the heights are the function values) so that you end up adding up different things if you don't change the function to compensate.
Best Answer
As per the author's work, we have:
$$f'(x) = a^{x} \lim_{h \to 0} \frac{a^h - 1}{a}$$
Replacing $x = 0$ in the above expression yields:
$$f'(0) = a^0 \lim_{h \to 0} \frac{a^h - 1}{a} = \lim_{h \to 0} \frac{a^h - 1}{a}$$
as desired.