Limits, Continuity & Differentiability
Squeeze Theorem — Strictly Increasing Function Limit
nta_pyq_2024_jan
Grade 12
Question:
Let $f:\mathbb{R}\to(0,\infty)$ be a strictly increasing function such that $\displaystyle\lim_{x\to\infty}\dfrac{f(7x)}{f(x)}=1$. Then, the value of $\displaystyle\lim_{x\to\infty}\left[\dfrac{f(5x)}{f(x)}-1\right]$ is equal to
Step-by-Step Solution
Key Concept: Since $f$ is strictly increasing: $f(x)<f(5x)<f(7x)$. Divide by $f(x)$: $1<\frac{f(5x)}{f(x)}<\frac{f(7x)}{f(x)}\to1$. By squeeze theorem, $\frac{f(5x)}{f(x)}\to1$, so $\left[\frac{f(5x)}{f(x)}-1\right]\to[0]=0$.
$f(x)<f(5x)<f(7x)\Rightarrow1<f(5x)/f(x)<f(7x)/f(x)\to1$. By squeeze, $f(5x)/f(x)\to1$. $\left[\frac{f(5x)}{f(x)}-1\right]\to[0]=0$.
Correct Answer: 2