alleryEvaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.
A$2$
B$4$
C$0$
D$1$
Answer & Solution
Correct answer: A. $2$
Use $1 - \cos 2x = 2\sin^2 x$. Then $\dfrac{2\sin^2 x}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2 \to 2 \cdot 1 = 2$ as $x \to 0$.
Alternatively, expand $\cos 2x \approx 1 - 2x^2$ to leading order, giving $\dfrac{2x^2}{x^2} = 2$.
Keep practising Mensuration: Surface Areas & Volumes
Browse all Mensuration: Surface Areas & Volumes questions for GRE, or explore more Mathematics topics.
Related questions
A line passes through the points $(1, 2, 3)$ and $(4, 6, 9)$. The direction ratios of the The direction cosines of a line are $l$, $m$, $n$. They satisfy:The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three-dimensiA fair coin is tossed twice. How many outcomes are there in the sample space?A hyperbola has foci at $(\pm 5, 0)$ and the length of the transverse axis is $6$. Find itWhat is the eccentricity of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$?For any hyperbola, the eccentricity $e$ satisfies:The standard equation of a hyperbola with centre at the origin and foci on the $x$-axis is