Limits and Squeeze Theorem
DAILY_CHALLENGE
Grade None
Question:
Let $f:(0,1)\to\mathbb{R}$ be the function defined as $f(x)=\sqrt{n}$ if $x\in\left[\dfrac{1}{n+1},\dfrac{1}{n}\right)$ where $n\in\mathbb{N}$. Let $g:(0,1)\to\mathbb{R}$ be a function such that $\displaystyle\int_{x^2}^{x}\sqrt{\dfrac{1-t}{t}}\,dt < g(x) < 2\sqrt{x}$ for all $x\in(0,1)$. Then $\displaystyle\lim_{x\to0^+}f(x)g(x)$
does NOT exist
is equal to 1
is equal to 2
is equal to 3
Step-by-Step Solution
Key Concept: Sandwich theorem: bound g between integral and 2√x; f(x)√x → 1 as x→0⁺
For $x\in[1/(n+1),1/n)$: $f(x)=\sqrt{n}$ and $x\approx1/n$ so $\sqrt{x}\approx1/\sqrt{n}$.
Upper bound: $f(x)\cdot2\sqrt{x}\leq\sqrt{n}\cdot2/\sqrt{n+1}\to2$ and $\geq\sqrt{n}\cdot2/\sqrt{n}=2$... actually more precisely $\sqrt{n}\cdot2\sqrt{x}$ with $x\in[1/(n+1),1/n)$ lies in $[2\sqrt{n/(n+1)},\,2)$. As $n\to\infty$: $2\sqrt{n/(n+1)}\to2$. So upper bound $\to2$.
Lower bound: $\int_{x^2}^{x}\sqrt{(1-t)/t}\,dt\approx\int_{x^2}^{x}(1/\sqrt{t})\,dt=2\sqrt{x}-2x$. So $f(x)\cdot(2\sqrt{x}-2x)\to2$ similarly.
By squeeze theorem: $\lim_{x\to0^+}f(x)g(x)=2$.
Correct Answer: C