x → 0 x (√(2 ² x + 3 x) - √( ² x + x + 4))$\lim_{x \to 0} \csc x \left(\sqrt{2\cos^2 x + 3\cos x} - \sqrt{\cos^2 x + \sin x + 4}\right)$ is equal to :

Related Formula

For a limit of 1∞$1^{\infty}$ form, x → a [g(x)]h(x)$\lim_{x \to a} [g(x)]^{h(x)}$ equals:

Core Logic

Let T = x → 1 ( (f(x + 2))/(f(3)) )(18)/((x - 1)²)$T = \lim_{x \to 1} \left( \frac{f(x + 2)}{f(3)} \right)^{\frac{18}{(x - 1)^2}}$. As x → 1$x \to 1$, (f(x+2))/(f(3)) → (f(3))/(f(3)) = 1$\frac{f(x+2)}{f(3)} \to \frac{f(3)}{f(3)} = 1$. The exponent goes to ∞$\infty$. This is a standard 1∞$1^{\infty}$ form.

Step 1: Simplify Exponent Limit

Given f(3) = 18$f(3) = 18$:

This is a (0)/(0)$\frac{0}{0}$ form limit.

Step 2: Apply L'Hopital's Rule

Differentiate numerator and denominator w.r.t x$x$:

This is still a (0)/(0)$\frac{0}{0}$ form since f'(3) = 0$f'(3) = 0$. Apply L'Hopital's Rule again:

Substitute x = 1$x = 1$:

Step 3: Final Calculation

Given f''(3) = 4$f''(3) = 4$:

The question asks for ₑ(T)$\log_e(T)$:

Pattern Recognition

When expanding f(x)$f(x)$ around an extrema (f'(a)=0$f'(a)=0$) inside a 1∞$1^{\infty}$ 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$f''(a)/2$.

Chapter Mix

Class 11 Maths: Limits and Derivatives Class 12 Maths: Limits, Continuity and Differentiability

Related Formula

Core Logic

Differentiate the given polynomial function iteratively to find expressions for f'(x)$f'(x)$ and f''(x)$f''(x)$, treating f'(1)$f'(1)$, f''(2)$f''(2)$, and f'''(3)$f'''(3)$ as constant values.

Step 1: First and Second Derivatives

Substitute x = 2$x = 2$ into the second derivative to create a relation:

Step 2: Substitute and Solve for Constants

Substitute f''(2)$f''(2)$ back into f'(x)$f'(x)$:

Now put x = 1$x = 1$ to solve for f'(1)$f'(1)$:

Step 3: Calculate Required Value

Determine the exact form of f'(x)$f'(x)$:

Substitute x = 5$x = 5$ to find f'(5)$f'(5)$:

Pattern Recognition

Treat derivatives evaluated at specific points (like f'(1)$f'(1)$) as fixed scalar constants. Substitute values back sequentially to solve the linear system of constants.

Chapter Mix

Class 12 Maths: Method of Differentiation

Related Formula

Core Logic

Evaluate f(x)$f(x)$ using Squeeze Theorem on the summation bounds due to the Greatest Integer Function.

Step 1: Evaluate Limit f(x)

Sum over bounds:

Divide by n³$n^3$ and apply limit n → ∞$n \to \infty$:

Step 2: Evaluate Final Summation

We need 12Σj=1∞ f(j)$12\sum_{j=1}^{\infty} f(j)$:

This is an infinite geometric progression with a = (1)/(9)$a = \frac{1}{9}$ and r = (1)/(3)$r = \frac{1}{3}$.

Finally, 12 × (1)/(6) = 2$12 \times \frac{1}{6} = 2$.

Pattern Recognition

When evaluating infinite limits over greatest integer sums n → ∞ 1np+1 Σ [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$0$.

Chapter Mix

Class 11 Maths: Limits, Continuity and Differentiability Class 11 Maths: Sequence and Series

Related Formula

Standard Taylor series expansions:

Core Logic

Expand numerator and denominator around x = 0$x = 0$: Denominator: x(1 - (x²)/(2)) - (x - (x²)/(2)) = (x²)/(2) + O(x³)$x\left(1 - \frac{x^2}{2}\right) - \left(x - \frac{x^2}{2}\right) = \frac{x^2}{2} + O(x^3)$. For limit to be finite, coefficients of x⁰$x^0$ and x¹$x^1$ in numerator must be zero:

• Coefficient of x⁰$x^0$: 1 + 2 + c - 2 = 0 c = -1$1 + 2 + c - 2 = 0 \implies c = -1$
• Coefficient of x¹$x^1$: (a-1) - (c-2) = 0 a - 1 + 3 = 0 a = -2$(a-1) - (c-2) = 0 \implies a - 1 + 3 = 0 \implies a = -2$

Step 1: Coefficient of x^2

Numerator coefficient of x²$x^2$ is ((a-1)²)/(2) - b² + (c-2)/(2)$\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2}$. Given limit value is 2$2$:

Step 2: Final Calculation

Pattern Recognition

Match powers of x$x$ in Taylor series to resolve indeterminate limit form (0)/(0)$\frac{0}{0}$.

Chapter Mix

Class 11 Maths: Limits, Continuity and Differentiability

Related Formula

Greatest integer function [x²]$[x^2]$ is discontinuous where x²$x^2$ takes integer values, except possibly where |x| = 0$|x| = 0$.

Core Logic

In (-2, 2)$(-2, 2)$, x² in [0, 4)$x^2 \in [0, 4)$. Integer values occur at x = 0, ± 1, ±√(2), ±√(3)$x = 0, \pm 1, \pm\sqrt{2}, \pm\sqrt{3}$. At x = 0$x = 0$, g(0) = 0$g(0) = 0$ and x → 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, -√(2), √(2), -√(3), √(3)$S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$.

Step 1: Evaluate f(x) for x in S

For f(x) = √(2)x, x²$f(x) = \min\{\sqrt{2}x, x^2\}$:

• f(-1) = -√(2), 1 = -√(2)$f(-1) = \min\{-\sqrt{2}, 1\} = -\sqrt{2}$
• f(1) = √(2), 1 = 1$f(1) = \min\{\sqrt{2}, 1\} = 1$
• f(-√(2)) = -2, 2 = -2$f(-\sqrt{2}) = \min\{-2, 2\} = -2$
• f(√(2)) = 2, 2 = 2$f(\sqrt{2}) = \min\{2, 2\} = 2$
• f(-√(3)) = -√(6), 3 = -√(6)$f(-\sqrt{3}) = \min\{-\sqrt{6}, 3\} = -\sqrt{6}$
• f(√(3)) = √(6), 3 = √(6)$f(\sqrt{3}) = \min\{\sqrt{6}, 3\} = \sqrt{6}$

Step 2: Summation

Pattern Recognition

Check origin continuity explicitly for |x|[x²]$|x|[x^2]$; evaluate √(2)x, x²$\min\{\sqrt{2}x, x^2\}$ case-by-case on set S$S$.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability