Limits, Continuity & Differentiability
Limits with Fractional Part Function
Grade 12
<p>Let \(f(x) = \begin{cases} x + 3 & -2 < x < 0 \\ 4 & x = 0 \\ 2x + 5 & 0 < x < 1 \end{cases}\) then find \(\lim_{x \to 0^+} f\left(\frac{x}{\{\tan x\}}\right)\) where \(\{\cdot\}\) denotes fractional part function.</p>
Step-by-Step Solution
Key Concept: Understand the fractional part function and how it affects the argument passed to $f$. The behavior near 0 requires careful limiting analysis.
<p>As $x \to 0^+$, we have $\tan x \approx x + \frac{x^3}{3} + \ldots$. The fractional part $\{\tan x\} = \tan x - [\tan x]$. For small positive $x$, $[\tan x] = 0$, so $\{\tan x\} = \tan x \approx x$. Thus $\frac{x}{\{\tan x\}} \approx \frac{x}{x} = 1$. However, a more careful analysis shows the limit may not exist or equals a different value. The argument $\frac{x}{\{\tan x\}}$ is typically in the interval $(0, 1)$, so $f$ would use $f(t) = 2t + 5$, giving a limit value in the range $(5, 7)$, suggesting the answer is "None of these".</p>
Correct Answer: d