Limit Calculation Using Taylor Expansion
We need to evaluate the limit:
$ L = \lim_{x \to 0} \frac{xtan 2x-2x tan x}{(1-cos2x)^2} $Substituting $x=0$ gives $\frac{0}{0}$, which is an indeterminate form. We can use Taylor series expansions around $x=0$.
Taylor Series Expansions
The relevant Taylor expansions are:
- $ \tan x = x + \frac{x^3}{3} + O(x^5) $
- $ \tan(2x) = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5) $
- $ \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + O(x^6) = 1 - 2x^2 + \frac{2x^4}{3} + O(x^6) $
Numerator Expansion
Let's expand the numerator $ N = xtan(2x) - 2x tan(x) $:
$ N = x \left( 2x + \frac{8x^3}{3} + O(x^5) \right) - 2x \left( x + \frac{x^3}{3} + O(x^5) \right) $ $ N = \left( 2x^2 + \frac{8x^4}{3} + O(x^6) \right) - \left( 2x^2 + \frac{2x^4}{3} + O(x^6) \right) $ $ N = \frac{8x^4}{3} - \frac{2x^4}{3} + O(x^6) = \frac{6x^4}{3} + O(x^6) = 2x^4 + O(x^6) $Denominator Expansion
Now, let's expand the denominator $ D = (1 - cos(2x))^2 $:
$ D = \left( 1 - \left( 1 - 2x^2 + \frac{2x^4}{3} + O(x^6) \right) \right)^2 $ $ D = \left( 2x^2 - \frac{2x^4}{3} + O(x^6) \right)^2 $The lowest order term dominates as $ x \to 0 $. So, $ D \approx (2x^2)^2 = 4x^4 $.
$ D = 4x^4 + O(x^6) $Evaluating the Limit
Substitute the simplified expansions back into the limit expression:
$ L = \lim_{x \to 0} \frac{2x^4 + O(x^6)}{4x^4 + O(x^6)} $Divide the numerator and denominator by $ x^4 $:
$ L = \lim_{x \to 0} \frac{2 + O(x^2)}{4 + O(x^2)} $As $ x \to 0 $, the higher order terms $ O(x^2) $ approach 0:
$ L = \frac{2}{4} = \frac{1}{2} $Conclusion
The value of the limit is $ \frac{1}{2} $. This corresponds to Option A.
Similar Questions
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_e x| - |x - 1|$:
(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then. - If the domain of the function $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
Let $[t]$ denote the greatest integer less than or equal to $t$. If the function
$f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases}$
is continuous at $x = 0$, then $a^2 + b^2$ is equal to
- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
- Let $f: \mathbf{R} \to \mathbf{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbf{R}$ and $f'(a-1) = 0$, where $a$ is a real number. Let $g(x) = f(\tan^2 x - 2\tan x + a)$, $0 < x < \frac{\pi}{2}$.
Consider the following two statements :
(I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$
(II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Then, - The sum of all the elements in the range of $f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x)$,
$x \neq \frac{n\pi}{2}, n \in \mathbf{Z}$, where $\text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \\ -1, & \text{if } t < 0 \end{cases}$, is : - The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is __________.
Let $f(x) = \begin{cases} (x-1)^\frac{1}{2-x}, & x > 1, x \ne 2 \\ k, & x = 2 \end{cases}$ The value of k for which f is continuous at x = 2 is :
- If $f(x) = \begin{cases} \frac{a|x| + x^2 - 2(\sin|x|)(\cos|x|)}{x} & , \ x \neq 0 \\ b & , \ x = 0 \end{cases}$ is continuous at $x = 0$, then $a + b$ is equal to
Important Questions from Differential Calculus
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_e x| - |x - 1|$:
(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then. - If the domain of the function $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x - 2} \right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
Let $[t]$ denote the greatest integer less than or equal to $t$. If the function
$f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases}$
is continuous at $x = 0$, then $a^2 + b^2$ is equal to
- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
