JEE-Mains 2015 (10/30)

Calculus Level 3

$\large \displaystyle \lim_{x \to 0} \dfrac{(1-\cos 2x)(3+\cos x)}{x \tan 4x}= \ ?$

\frac{1}{2}

4 2

1 solution

Chew-Seong Cheong
Jun 14, 2015

$\begin{aligned} \lim_{x \to 0} \frac{(1-\cos{2x})(3+\cos{x})}{x\tan{4x}} & = \lim_{x \to 0} \frac{(1-[1-2\sin^2{x}])(3+\cos{x})\cos{4x}}{x\sin{4x}} \\ & = \lim_{x \to 0} \frac{2\sin^2{x}(3+\cos{x})\cos{4x}}{2x\sin{2x}\cos{2x}} \\ & = \lim_{x \to 0} \frac{\sin^2{x}(3+\cos{x})\cos{4x}}{2x\sin{x}\cos{x}\cos{2x}} \\ & = \lim_{x \to 0} \left(\frac{\sin{x}}{x}\right) \frac{(3+\cos{x})\cos{4x}}{2\cos{x}\cos{2x}} \\ & = (1) \frac{(3+1)(1)}{2(1)(1)} = \boxed{2} \end{aligned}$

$\color{#D61F06}{\text{In response to Challenge Master:}}$

$\begin{aligned} \lim_{x \to 0} \frac{(1-\cos{2x})(3+\cos{x})}{x\tan{4x}} & = \lim_{x \to 0} \frac{(1-1 + 2\sin^2{x})(3+\cos{x})}{x\tan{4x}} \\ & = \lim_{x \to 0} \frac{(2\sin^2{x})(3+\cos{x})}{x\tan{4x}} \\ & = \frac{(2x^2)(3+1)}{x(4x)} = \boxed{2} \end{aligned}$

Moderator note:

You can use the approximations $\tan(x) \approx x$ and $\cos(x) \approx 1 - \frac{x^2}2$ for small $x$ . Do you see why?

Yes, $\lim_{x \to 0} \sin{x} = x$ . Lack of practice and failed to see it.

Chew-Seong Cheong - 5 years, 11 months ago

Why sinx/x equals to 1 when the limit approaches zero? If it is 0/0, then the answer is NaN.

Tapas Mazumdar - 5 years, 11 months ago

$\lim_{x \to 0} \sin{x} = x$ , therefore, $\lim_{x \to 0} \dfrac{\sin{x}}{x} = 1$

Chew-Seong Cheong - 5 years, 11 months ago

JEE-Mains 2015 (10/30)

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