Limits, Continuity & Differentiability
Limits with Piecewise Functions
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 \(\lim_{x \to 0^-} f([x - \tan x])\) where \([\cdot]\) denotes greatest integer function.</p>
Step-by-Step Solution
Key Concept: Carefully analyze the behavior of $x - \tan x$ near $x = 0^-$ and determine which integer value the greatest integer function yields.
<p>As $x \to 0^-$, we have $\tan x \to 0^-$ and $x \to 0^-$. Therefore $x - \tan x \to 0$. More precisely, using Taylor expansion: $\tan x = x + \frac{x^3}{3} + \ldots$, so $x - \tan x = -\frac{x^3}{3} - \ldots \to 0^-$. Thus $[x - \tan x] = -1$ for $x$ close to $0^-$. Since $-2 < -1 < 0$, we have $f(-1) = -1 + 3 = 2$. Wait, rechecking: $[x - \tan x]$ approaches $-1$, so $\lim_{x \to 0^-} f([x - \tan x]) = f(-1) = -1 + 3 = 2$. However, if the greatest integer of the argument is 0, then $f(0) = 4$.</p>
Correct Answer: b