Q.Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous: $f(x) = \dfrac{x^3-8}{x^2-4}$, for $x > 2$, $= 3$, for $x = 2$, $= \dfrac{e^{3(x-2)^2}-1}{2(x-2)^2}$, for $x < 2$.

$f(x)=\dfrac{x^3-8}{x^2-4}$ for $x>2$, $f(2)=3$, and $f(x)=\dfrac{e^{3(x-2)^2}-1}{2(x-2)^2}$ for $x<2$.

Right-hand limit: factor $x^3-8=(x-2)(x^2+2x+4)$ and $x^2-4=(x-2)(x+2)$, so for $x\ne2$, $\dfrac{x^3-8}{x^2-4}=\dfrac{x^2+2x+4}{x+2}$, giving $\displaystyle\lim_{x\to2^+} f(x)=\dfrac{4+4+4}{4}=\dfrac{12}{4}=3$.

Left-hand limit: put $h=(x-2)^2\to0^+$ as $x\to2^-$; then $\displaystyle\lim_{x\to2^-} f(x)=\lim_{h\to0^+}\frac{e^{3h}-1}{2h}=\frac12\lim_{h\to0}\frac{e^{3h}-1}{h}=\frac12(3)=\frac32$, using $\lim_{h\to0}(e^{3h}-1)/h=3$. …

Simplify each branch separately near $x=2$ — factor the rational branch for $x>2$, and treat $(x-2)^2$ as a single small variable for the expone …

Stopping after checking only that the right-hand limit matches $f(2)=3$ and concluding the function is continuous — the left- …