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
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- 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 ____________.
Important Questions from Differential Calculus
- The number of points of discontinuity of the function $f (x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}], x \in [0,4]$, where $[.]$denotes the greatest integer function, is ________
Let the function $f (x) = \frac{x}{3} + \frac{3}{x} + 3$, $x\neq0$ be strictly increasing in $(-\infty, a_1)\cup(a_2,\infty)$ and strictly decreasing in $(a_3, \alpha_4)\cup(a_4,a_5)$. Then $\sum_{i=1}^5 a_i^2$ is equal to
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- 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, - If the domain of the function $f(x) = \sin^{-1}\left(\frac{5-x}{3+2x}\right) + \frac{1}{\log_e(10-x)}$ is $(-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}$, then $6(\alpha + \beta + \gamma + \delta)$ is equal to
