First slide

Introduction to limits

Question

Let $p = lim_{x \to 0} \frac{\ln (\cos 2 x)}{3 x^{2}}, q = lim_{x \to 0} \frac{\sin^{2} 2 x}{x (1 - e^{x})}$ and $r = lim_{x \to 1} \frac{\sqrt{x} - x}{\ln x}$ Then p, e, r satisfy

Moderate

A

p<q<r
B

q<r<p
C

p<r<q
D

q<p<r

Solution

Clearly,
$\begin{array}{l} p = lim_{x \to 0} \frac{\ln (1 + \cos 2 x - 1)}{3 x^{2}} \\ = lim_{x \to 0} \frac{\ln (1 + \cos 2 x - 1)}{(\cos 2 x - 1)} \cdot \frac{\cos 2 x - 1}{3 x^{2}} = - \frac{2}{3} \\ q = lim_{x \to 0} \frac{\sin^{2} 2 x}{4 x^{2}} \cdot \frac{4 x^{2}}{x (1 - e^{x})} = - 4 \\ and r = lim_{x \to 1} \frac{\sqrt{x} - x}{\ln (1 + x - 1)} = lim_{x \to 1} \frac{\sqrt{x} (1 - \sqrt{x})}{\ln (\frac{1 + x - 1}{x - 1}) \cdot (x - 1)} \\ = lim_{x \to 1} \frac{\sqrt{x} (1 - x)}{\ln (\frac{1 + (x - 1)}{x - 1}) \cdot (x - 1) (1 + \sqrt{x})} = - \frac{1}{2} \end{array}$
Hence, q < p < r

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