2018 international practice exam\n10 mark for review\nthe function f is continuous and increasing for x≥…
2018 international practice exam\n10 mark for review\nthe function f is continuous and increasing for x≥ - 1. the table above gives values of f at selected values of x.\nof the following, which is the best approximation for lim e^(-2f(x))?\na -2\nb 0\nc 1\nd the limit does not exist
Answer
Explanation:
Step1: Analyze the limit of f(x) as x→0
As x approaches 0 from the left (values like -0.1, -0.01, -0.001), f(x) approaches 0. As x approaches 0 from the right (values like 0.001, 0.01, 0.1), f(x) also approaches 0. Since f(x) is continuous, $\lim_{x\rightarrow0}f(x)=0$.
Step2: Use the limit - rule for exponential function
We want to find $\lim_{x\rightarrow0}e^{- 2f(x)}$. Let $y = - 2f(x)$. Then $\lim_{x\rightarrow0}y=-2\lim_{x\rightarrow0}f(x)$. Substituting $\lim_{x\rightarrow0}f(x) = 0$, we get $\lim_{x\rightarrow0}y = 0$.
Step3: Find the limit of the exponential function
We know that the exponential function $y = e^{u}$ is continuous. So $\lim_{x\rightarrow0}e^{-2f(x)}=e^{\lim_{x\rightarrow0}(-2f(x))}$. Since $\lim_{x\rightarrow0}(-2f(x)) = 0$, then $\lim_{x\rightarrow0}e^{-2f(x)}=e^{0}=1$.
Answer:
C. 1