Using a limit to find the slope of the graph of $\log_k x$ at $(1,0)$

Question

I have the following problem:

Consider the curve $f(x) = \log_k(x)$. Show that the slope of $f(x)$ at $(1,0)$ is given by $\lim_{h \to 0} \frac{\log_k(1+h)}{h}$ and, using properties of logarithms and the substitution $n=\frac{1}{h}$, show that this limit is $\log_ke$.

I just don't quite understand. I don't see a place to apply logarithm properties in a meaningful way. You can do $\log_k(1+h) = \frac{\log(1+h)}{\log(k)}$, but I don't see what that achieves. I also have no idea where you could apply the mentioned substitution.

Any ideas? Thanks.

Answer 1

From the definition of the derivative, we have that

$$f'(x) = \lim_{h \to 0} \frac{\log_k (x+h) - \log_k(x)}{h}$$

Then $f'(1) = \lim_{h \to 0} \frac{\log_k (1+h) - \log_k(1)}{h}$, and since $\log_k 1 = 0$ for any base $k$, we have that $f'(1) = \lim_{h \to 0} \frac{\log_k (1+h)}{h}$.

Now, let $y = \lim_{h \to 0} \frac{\log_k (1+h)}{h}$. Then $e^y = \lim_{h \to 0}e^{\frac{\log_k (1+h)}{h}} = \lim_{h \to 0}e^{{\log_k (1+h)}^{\frac1h}}$. Put $n = \frac 1h$. Then as $h \to 0$, $n \to \pm \infty$. Our new limit is $\lim_{n \to \infty} e^{\log_k (1\pm\frac 1n)^\pm n} = e^{\log_k e}$.

Now we have $e^y = e^{\log_k e}$, so that $y = \log_k e$ as desired.

Answer 2

\begin{align} f'(x) & = \lim_{h\to0} \frac{f(x+h) - f(x)} h = \lim_{h\to0} \frac{\log_k(1+h) - \log_k 1} h = \lim_{h\to0} \frac{\log_k(1+h)} h \\[10pt] & = \lim_{h\to 0} \frac{\left( \frac{\log(1+h)}{\log k} \right)} h = \frac 1 {\log k} \lim_{h\to0} \frac{\log(1+h)} h = \frac 1 {\log k} \lim_{h\to0} \frac{\log(1+h) - \log 1} h \\[10pt] & = \frac 1 {\log k} \cdot \left. \frac d {dx} \log x\, \right|_{x=1} = \frac 1 {\log_e k} = \log_k e. \end{align}

Answer 3

$$\begin{align} \frac d{dx}\log_k(x)&=\lim_{h\to0} \frac{\log_k(x+h)-\log_k(x)}h \\ &=\lim_{h\to0} \frac{\log_k\left(\frac{x+h}x\right)}h \\ &=\lim_{h\to0} \frac{\log_k\left(1+\frac hx\right)}h \\ &=\lim_{h\to0} \frac{\log_k(1+h)}h \\~\\ &=\lim_{h\to0} \frac1h\log_k(1+h) \\ &=\lim_{h\to0} \log_k\left((1+h)^{1/h}\right) \\ &=\lim_{n\to\infty} \log_k\left(\left(1+\frac1n\right)^n\right) \\ &=\log_ke \end{align}$$

Answer 4

we have by the definition of the slope $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{\log_k (1+h)-\log_k (1)}{h}=\frac{\log_k (1+h)}{h}$ . Now $\lim_{h \to 0}\frac{\log_k (1+h)}{h} = \lim_{h \to 0}\log_k (1+h)^{(1/h)}=\lim_{n \to +\infty}\log_k (1+1/n)^{n}=\log_ke$.