MA25C01 | Calculus 🎯|Example-6💡 Piecewise Functions: Evaluating Limits & Continuity (Step-by-Step) |

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[N/D-23-R-21]

Example

Let \(f(x)=\begin{cases} \sqrt{-x}, & x<0 \\ 3-x, & 0 \leq x \leq 3 \\ (x-3)^2, & x>3 \end{cases}\)
Evaluate the following limits, if they exist:

(i) \(\lim_{x\to 0^{-}} f(x)\)	(ii) \(\lim_{x\to 0^{+}} f(x)\)	(iii) \(\lim_{x\to 3^{-}} f(x)\)
(iv) \(\lim_{x\to 3^{+}} f(x)\)	(v) \(\lim_{x\to 0} f(x)\)	(vi) \(\lim_{x\to 3} f(x)\)

Also find where \(f(x)\) is continuous.

Solution

\begin{align*} (i) & \lim_{x\to 0^{-}} f(x) = \lim_{x\to 0} \sqrt{-x} = 0 & (ii) & \lim_{x\to 0^{+}} f(x) = \lim_{x\to 0} (3-x) = 3 \\ \\ (iii) & \lim_{x\to 3^{-}} f(x) = \lim_{x\to 3} (3-x) = 0 & (iv) & \lim_{x\to 3^{+}} f(x) = \lim_{x\to 3} (x-3)^2 = 0 \\ \\ (v) & \lim_{x\to 0} f(x) \text{ does not exist because } \lim_{x\to 0^{-}} f(x) \neq \lim_{x\to 0^{+}} f(x) \\ (vi) & \lim_{x\to 3} f(x) = 0 \text{ because } \lim_{x\to 3^{-}} f(x) = \lim_{x\to 3^{+}} f(x) = 0 \end{align*}

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