AP DSC School Assistant Math Model Test 3

Concept:The limit is of the indeterminate form $\frac{0}{0}$, so L'Hôpital's rule (differentiating numerator and denominator separately) can be applied repeatedly until the limit is determined.Explanation:1. First, substitute $x=0$: numerator $e^0 + e^0 - 2 = 1+1-2=0$, denominator $0^2=0$. Hence it is $\frac{0}{0}$ form.2. Apply L'Hôpital's rule: differentiate numerator and denominator. $\frac{d}{dx}(e^x+e^{-x}-2) = e^x - e^{-x}$, $\frac{d}{dx}(x^2)=2x$. So the limit becomes $\displaystyle\lim_{x\to 0}\frac{e^x - e^{-x}}{2x}$, which is again $\frac{0}{0}$ because $e^0-e^0=0$ and $2\cdot0=0$.3. Apply L'Hôpital's rule a second time: differentiate again. $\frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x}$, $\frac{d}{dx}(2x)=2$. So the limit becomes $\displaystyle\lim_{x\to 0}\frac{e^x+e^{-x}}{2}$.4. Now substitute $x=0$: $e^0+e^0 = 1+1 = 2$, so $\frac{2}{2}=1$.Answer:The correct answer is $1$ (option C).