Q.Find $k$ if the following function is continuous at the point indicated against it: $f(x) = \left(\dfrac{5x-8}{8-3x}\right)^{\frac{3}{2x-4}}$, for $x \ne 2$, $= k$, for $x=2$, at $x=2$.

$f(x)=\left(\dfrac{5x-8}{8-3x}\right)^{\frac{3}{2x-4}}$ for $x\ne2$, $f(2)=k$, continuous at $2$.

At $x=2$: base $=\dfrac{10-8}{8-6}=1$, while the exponent $\dfrac{3}{2x-4}=\dfrac{3}{2(x-2)}\to\infty$ — a $1^\infty$ form. Put $x=2+h$, $h\to0$: base $=\dfrac{2+5h}{2-3h}=1+\dfrac{(2+5h)-(2-3h)}{2-3h}=1+\dfrac{8h}{2-3h}$, so $g(h)=\dfrac{8h}{2-3h}\to0$ as $h\to0$. …

Substitute $x=2+h$ to turn the $1^\infty$ form into $\big(1+g(h)\big)^{3/(2h)}$ with $g(h)=\dfrac{8h}{2-3h}\to0 …

Simplifying the base's numerator difference incorrectly (e.g. writing $g(h)$ as $4h/(2-3h)$ instead of the correct $8h/(2-3h)$) — recompute $(2+5h)-(2-3h)=8h$ directly rather …