Let m & n be two positive numbers greater than 1 . If lim_(p→0) ((e^(cos (p^n )) −e)/p^m ) = (1/2)e then (n/m)=?


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 162399 by cortano last updated on 29/Dec/21

$L e t m & n b e t w o p o s i t i v e n u m b e r s$ $g r e a t e r t h a n 1 . I f \lim_{p \to 0} \frac{e^{\cos (p^{n})} - e}{p^{m}} = \frac{1}{2} e$ $t h e n \frac{n}{m} = ?$
Commented byblackmamba last updated on 29/Dec/21

$\lim_{p \to 0} \frac{e^{\cos (p^{n})} - e}{p^{m}} = e \lim_{p \to 0} \frac{e^{\cos (p^{n}) - 1} - 1}{p^{m}} = \frac{1}{2} e$ $\lim_{p \to 0} \frac{e^{\cos (p^{n}) - 1} - 1}{\cos (p^{n}) - 1} \times \lim_{p \to 0} \frac{\cos (p^{n}) - 1}{p^{m}} = \frac{1}{2}$ $[n o t i c e t h a t \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1]$ $[\lim_{p \to 0} \frac{e^{\cos (p^{n}) - 1} - 1}{\cos (p^{n}) - 1} = 1]$ $\Leftrightarrow \lim_{p \to 0} \frac{\cos (p^{n}) - 1}{p^{m}} = \frac{1}{2}$ $\Leftrightarrow \lim_{p \to 0} \frac{- 2 \sin^{2} (\frac{p^{n}}{2})}{p^{m}} = \frac{1}{2}$ $\Leftrightarrow \lim_{p \to 0} \frac{{(\frac{\sin (\frac{p^{n}}{2})}{(\frac{p^{n}}{2})})}^{2} . (\frac{p^{2 n}}{4})}{p^{m}} \overset{?}{=} - \frac{1}{4}$ $2 n = m \Rightarrow \frac{n}{m} = \frac{n}{2 m} = \frac{1}{2}$
	Answered by qaz last updated on 29/Dec/21

	$\lim_{p \to 0} \frac{e^{\cos (p^{n})} - e}{p^{m}}$ $= \lim_{p \to 0} \frac{e (e^{\cos (p^{n}) - 1} - 1)}{p^{m}}$ $= e \lim_{p \to 0} \frac{\cos (p^{n}) - 1}{p^{m}}$ $= e \lim_{p \to 0} \frac{- \frac{1}{2} p^{2 n}}{p^{m}}$ $= - \frac{1}{2} e$ $\Rightarrow \frac{n}{m} = \frac{1}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum