Q.Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous: $f(x) = \log_{(1+3x)}(1+5x)$ for $x > 0$, $= \dfrac{32^x-1}{8^x-1}$, for $x < 0$, at $x = 0$.

$f(x)=\log_{(1+3x)}(1+5x)$ for $x>0$, and $f(x)=\dfrac{32^x-1}{8^x-1}$ for $x<0$; $f(0)$ is not assigned by either strict inequality.

Right-hand limit: $\log_{(1+3x)}(1+5x)=\dfrac{\ln(1+5x)}{\ln(1+3x)}=\dfrac{\ln(1+5x)/x}{\ln(1+3x)/x}\to\dfrac{5}{3}$ as $x\to0^+$, using $\lim_{x\to0}\ln(1+kx)/x=k$.

Left-hand limit: $\dfrac{32^x-1}{8^x-1}=\dfrac{(32^x-1)/x}{(8^x-1)/x}\to\dfrac{\ln32}{\ln8}=\dfrac{5\ln2}{3\ln2}=\dfrac53$ as $x\to0^-$, using $\lim_{x\to0}(a^x-1)/x=\ln a$. …

Convert the change-of-base logarithm to natural logs, and rewrite $32^x-1$ and $8^x-1$ as their own standard $(a^x-1)/x\to\ln a$ limit …

Forgetting that $\ln32=5\ln2$ and $\ln8=3\ln2$ share the common factor $\ln2$, which cancels — without spotting this, the ratio $\ln32/ …