If lim_x to 0 frace^(a-1)x + 2 cos bx + (c-2)e^-xx cos x - log_e(1+x) = 2$\lim_{x \to 0} \frac{e^{(a-1)x} + 2 \cos bx + (c-2)e^{-x}}{x \cos x - \log_e(1+x)} = 2$, then a^2 + b^2 + c^2$a^2 + b^2 + c^2$ is equal to:

### Related Formula For a limit of 1^infty$1^{\infty}$ form, lim_x to a [g(x)]^h(x)$\lim_{x \to a} [g(x)]^{h(x)}$ equals: e^lim_x to a h(x)[g(x) - 1]$$e^{\lim_{x \to a} h(x)[g(x) - 1]}$$ ### Core Logic Let T = lim_x to 1 left( fracf(x + 2)f(3) right)^frac18(x - 1)^2$T = \lim_{x \to 1} \left( \frac{f(x + 2)}{f(3)} \right)^{\frac{18}{(x - 1)^2}}$. As x to 1$x \to 1$, fracf(x+2)f(3) to fracf(3)f(3) = 1$\frac{f(x+2)}{f(3)} \to \frac{f(3)}{f(3)} = 1$. The exponent goes to infty$\infty$. This is a standard 1^infty$1^{\infty}$ form. T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)f(3) right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{f(3)} \right)}$$ ### Step 1: Simplify Exponent Limit Given f(3) = 18$f(3) = 18$: T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)18 right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{18} \right)}$$ T = e^lim_x to 1 fracf(x + 2) - f(3)(x - 1)^2$$T = e^{\lim_{x \to 1} \frac{f(x + 2) - f(3)}{(x - 1)^2}}$$ This is a frac00$\frac{0}{0}$ form limit. ### Step 2: Apply L'Hopital's Rule Differentiate numerator and denominator w.r.t x$x$: T = e^lim_x to 1 fracf'(x + 2)2(x - 1)$$T = e^{\lim_{x \to 1} \frac{f'(x + 2)}{2(x - 1)}}$$ This is still a frac00$\frac{0}{0}$ form since f'(3) = 0$f'(3) = 0$. Apply L'Hopital's Rule again: T = e^lim_x to 1 fracf''(x + 2)2$$T = e^{\lim_{x \to 1} \frac{f''(x + 2)}{2}}$$ Substitute x = 1$x = 1$: T = e^fracf''(3)2$$T = e^{\frac{f''(3)}{2}}$$ ### Step 3: Final Calculation Given f''(3) = 4$f''(3) = 4$: T = e^frac42 = e^2$$T = e^{\frac{4}{2}} = e^2$$ The question asks for log_e(T)$\log_e(T)$: log_e(T) = log_e(e^2) = 2$$\log_e(T) = \log_e(e^2) = 2$$ ### Pattern Recognition When expanding f(x)$f(x)$ around an extrema (f'(a)=0$f'(a)=0$) inside a 1^infty$1^{\infty}$ limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (2! = 2$2! = 2$). The limit elegantly shrinks directly to f''(a)/2$f''(a)/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Limits, Continuity and Differentiability

### Related Formula textSandwich Theorem for greatest integer function: x - 1 < [x] leq x$$\text{Sandwich Theorem for greatest integer function: } x - 1 < [x] \leq x$$ sum_k=1^n k^2 = fracn(n+1)(2n+1)6$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ ### Core Logic Evaluate f(x)$f(x)$ using Squeeze Theorem on the summation bounds due to the Greatest Integer Function. frack^23^x - 1 < left[frack^23^xright] leq frack^23^x$$\frac{k^2}{3^x} - 1 < \left[\frac{k^2}{3^x}\right] \leq \frac{k^2}{3^x}$$ ### Step 1: Evaluate Limit f(x) Sum over bounds: sum_k=1^n left( frack^23^x - 1 right) < sum_k=1^n left[ frack^23^x right] leq sum_k=1^n frack^23^x$$\sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \sum_{k=1}^n \frac{k^2}{3^x}$$ frac13^x fracn(n+1)(2n+1)6 - n < sum_k=1^n left[ frack^23^x right] leq frac13^x fracn(n+1)(2n+1)6$$\frac{1}{3^x} \frac{n(n+1)(2n+1)}{6} - n < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \frac{1}{3^x} \frac{n(n+1)(2n+1)}{6}$$ Divide by n^3$n^3$ and apply limit n to infty$n \to \infty$: lim_n to infty left( frac2n^3 + 3n^2 + n6n^3 cdot 3^x - frac1n^2 right) < f(x) leq lim_n to infty frac2n^3 + 3n^2 + n6n^3 cdot 3^x$$\lim_{n \to \infty} \left( \frac{2n^3 + 3n^2 + n}{6n^3 \cdot 3^x} - \frac{1}{n^2} \right) < f(x) \leq \lim_{n \to \infty} \frac{2n^3 + 3n^2 + n}{6n^3 \cdot 3^x}$$ f(x) = frac26 cdot 3^x = frac13 cdot 3^x = frac13^x+1$$f(x) = \frac{2}{6 \cdot 3^x} = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$$ ### Step 2: Evaluate Final Summation We need 12sum_j=1^infty f(j)$12\sum_{j=1}^{\infty} f(j)$: 12 sum_j=1^infty frac13^j+1 = 12 left( frac13^2 + frac13^3 + ldots right)$$12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = 12 \left( \frac{1}{3^2} + \frac{1}{3^3} + \ldots \right)$$ This is an infinite geometric progression with a = frac19$a = \frac{1}{9}$ and r = frac13$r = \frac{1}{3}$. textSum = fraca1 - r = frac1/91 - 1/3 = frac1/92/3 = frac16$$\text{Sum} = \frac{a}{1 - r} = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{6}$$ Finally, 12 times frac16 = 2$12 \times \frac{1}{6} = 2$. ### Pattern Recognition When evaluating infinite limits over greatest integer sums lim_n to infty frac1n^p+1 sum [k^p/C]$\lim_{n \to \infty} \frac{1}{n^{p+1}} \sum [k^p/C]$, the G.I.F brackets can be safely replaced by exact values due to Sandwich theorem bounding the -1$-1$ residual to 0$0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits, Continuity and Differentiability Class 11 Maths: Sequence and Series

### Related Formula Greatest integer function [x^2]$[x^2]$ is discontinuous where x^2$x^2$ takes integer values, except possibly where |x| = 0$|x| = 0$. ### Core Logic In (-2, 2)$(-2, 2)$, x^2 in [0, 4)$x^2 \in [0, 4)$. Integer values occur at x = 0, pm 1, pmsqrt2, pmsqrt3$x = 0, \pm 1, \pm\sqrt{2}, \pm\sqrt{3}$. At x = 0$x = 0$, g(0) = 0$g(0) = 0$ and lim_x to 0 g(x) = 0$\lim_{x \to 0} g(x) = 0$, so g(x)$g(x)$ is continuous at x = 0$x = 0$. Points of discontinuity: S = \-1, 1, -sqrt2, sqrt2, -sqrt3, sqrt3\$S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$. ### Step 1: Evaluate f(x) for x in S For f(x) = min\sqrt2x, x^2\$f(x) = \min\{\sqrt{2}x, x^2\}$: - f(-1) = min\-sqrt2, 1\ = -sqrt2$f(-1) = \min\{-\sqrt{2}, 1\} = -\sqrt{2}$ - f(1) = min\sqrt2, 1\ = 1$f(1) = \min\{\sqrt{2}, 1\} = 1$ - f(-sqrt2) = min\-2, 2\ = -2$f(-\sqrt{2}) = \min\{-2, 2\} = -2$ - f(sqrt2) = min\2, 2\ = 2$f(\sqrt{2}) = \min\{2, 2\} = 2$ - f(-sqrt3) = min\-sqrt6, 3\ = -sqrt6$f(-\sqrt{3}) = \min\{-\sqrt{6}, 3\} = -\sqrt{6}$ - f(sqrt3) = min\sqrt6, 3\ = sqrt6$f(\sqrt{3}) = \min\{\sqrt{6}, 3\} = \sqrt{6}$ ### Step 2: Summation sum_x in S f(x) = -sqrt2 + 1 - 2 + 2 - sqrt6 + sqrt6 = 1 - sqrt2$$\sum_{x \in S} f(x) = -\sqrt{2} + 1 - 2 + 2 - \sqrt{6} + \sqrt{6} = 1 - \sqrt{2}$$ ### Pattern Recognition Check origin continuity explicitly for |x|[x^2]$|x|[x^2]$; evaluate min\sqrt2x, x^2\$\min\{\sqrt{2}x, x^2\}$ case-by-case on set S$S$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Limits, Continuity and Differentiability

### Related Formula textTaylor Series expansion of cos u = 1 - fracu^22 + fracu^424 + O(u^6)$$\text{Taylor Series expansion of } \cos u = 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6)$$ ### Core Logic Since the denominator has x^4$x^4$, we expand the numerator using Taylor series up to x^4$x^4$. For the limit to exist and be finite, the coefficients of lower powers of x$x$ (specifically x^0$x^0$ and x^2$x^2$) must be zero. ### Step 1: Write down series expansions Expand cos(2x)$\cos(2x)$ and cos(4x)$\cos(4x)$: cos 2x = 1 - frac4x^22 + frac16x^424 + O(x^6) = 1 - 2x^2 + frac23x^4 + O(x^6)$$\cos 2x = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} + O(x^6) = 1 - 2x^2 + \frac{2}{3}x^4 + O(x^6)$$ cos 4x = 1 - frac16x^22 + frac256x^424 + O(x^6) = 1 - 8x^2 + frac323x^4 + O(x^6)$$\cos 4x = 1 - \frac{16x^2}{2} + \frac{256x^4}{24} + O(x^6) = 1 - 8x^2 + \frac{32}{3}x^4 + O(x^6)$$ ### Step 2: Collect coefficients in the numerator The numerator of the limit is: cos(2x) + acos(4x) - b = left( 1 - 2x^2 + frac23x^4 right) + aleft( 1 - 8x^2 + frac323x^4 right) - b$$\cos(2x) + a\cos(4x) - b = \left( 1 - 2x^2 + \frac{2}{3}x^4 \right) + a\left( 1 - 8x^2 + \frac{32}{3}x^4 \right) - b$$ = (1 + a - b) - x^2(2 + 8a) + x^4left(frac23 + frac323aright) + O(x^6)$$= (1 + a - b) - x^2(2 + 8a) + x^4\left(\frac{2}{3} + \frac{32}{3}a\right) + O(x^6)$$ ### Step 3: Set lower order coefficients to zero For the limit to be finite, the coefficients of x^0$x^0$ and x^2$x^2$ must vanish: - From x^2$x^2$ coefficient: 2 + 8a = 0 implies a = -frac14$$2 + 8a = 0 \implies a = -\frac{1}{4}$$ - From constant term: 1 + a - b = 0 implies b = a + 1 = -frac14 + 1 = frac34$$1 + a - b = 0 \implies b = a + 1 = -\frac{1}{4} + 1 = \frac{3}{4}$$ Now calculate the sum: a + b = -frac14 + frac34 = frac12$$a + b = -\frac{1}{4} + \frac{3}{4} = \frac{1}{2}$$ ### Pattern Recognition Finiteness condition: When a limit is finite with a denominator of x^n$x^n$, it implies that the numerator is a function of order O(x^n)$O(x^n)$ near zero. Taylor expansions allow you to quickly extract the necessary values of unknown coefficients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability