### Related Formula Continuity definition condition frame: lim_x rightarrow 0^- f(x) = f(0) = lim_x rightarrow 0^+ f(x)$\lim_{x \rightarrow 0^-} f(x) = f(0) = \lim_{x \rightarrow 0^+} f(x)$ ### Core Logic Evaluate Left-Hand Limit (LHL) using standard forms [cite: 1410]: textLHL = lim_x rightarrow 0^- (1+ax)^1/x = e^a$\text{LHL} = \lim_{x \rightarrow 0^-} (1+ax)^{1/x} = e^a$ [cite: 1410] Given baseline definition states f(0) = 1+b$f(0) = 1+b$ [cite: 1410]. For Right-Hand Limit (RHL) to be finite and valid, the numerator tracking towards 0 means the denominator must also balance towards 0 to avoid divergence [cite: 1411]: lim_x rightarrow 0^+ left[(x+c)^1/3 - 2right] = 0 implies c^1/3 = 2 implies c = 8$\lim_{x \rightarrow 0^+} \left[(x+c)^{1/3} - 2\right] = 0 \implies c^{1/3} = 2 \implies c = 8$ [cite: 1414] ### Step 1: Applying L'Hopital's rule to the RHL With c=8$c=8$, evaluate RHL limit expressions using derivatives [cite: 1411]: textRHL = lim_x rightarrow 0^+ fracfrac12sqrtx+4frac13(x+8)^-2/3 = fracfrac12(2)frac13(8)^-2/3 = fracfrac14frac13 cdot 4 = frac14 cdot 12 = 3$\text{RHL} = \lim_{x \rightarrow 0^+} \frac{\frac{1}{2\sqrt{x+4}}}{\frac{1}{3}(x+8)^{-2/3}} = \frac{\frac{1}{2(2)}}{\frac{1}{3}(8)^{-2/3}} = \frac{\frac{1}{4}}{\frac{1}{3 \cdot 4}} = \frac{1}{4} \cdot 12 = 3$ [cite: 1411, 1415] ### Step 2: Equating limits for parameter solutions Equate continuous criteria milestones together [cite: 1416]: e^a = 1 + b = 3$e^a = 1 + b = 3$ [cite: 1416] e^a = 3 implies b = 2$e^a = 3 \implies b = 2$ [cite: 1416] Compute the ultimate target combination product configuration [cite: 1416]: e^a cdot b cdot c = 3 cdot 2 cdot 8 = 48$e^a \cdot b \cdot c = 3 \cdot 2 \cdot 8 = 48$ [cite: 1416] ### Pattern Recognition Determining missing root constants inside indeterminate fraction structures handles calculations swiftly before running formal limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability