Let-m-amp-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 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 by blackmamba 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}

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