Continuity practice question 25

What sign does the exponent $1/x$ have just left of 0, and just right of it?

Treat the two sides separately. As $x \to 0^{-}$, $1/x \to -\infty$; as $x \to 0^{+}$, $1/x \to +\infty$. What does $e^{1/x}$ do in each case?

From the right, $e^{1/x}$ blows up in both the numerator and the denominator. Divide both by it: $f(x) = \dfrac{5 + 2e^{-1/x}}{1 + e^{-1/x}}$.