L-lim-x-0-e-x-cos-x-x-1-x-3-

Question Number 141900 by iloveisrael last updated on 24/May/21

L = \lim_{x \to 0} \frac{e^{x} \cos x - x - 1}{x^{3}} = ?

Answered by Dwaipayan Shikari last updated on 24/May/21

\lim_{x \to 0} \frac{e^{x} (c o s x) - x - 1}{x^{3}} = \frac{e^{x} (1 - \frac{x^{2}}{2}) - x - 1}{x^{3}}

= \frac{e^{x} - x - 1 - e^{x} \frac{x^{2}}{2}}{x^{3}} = \frac{\frac{x^{2}}{2} + \frac{x^{3}}{3!} + O (x^{4}) - (1 + x + O (x^{2})) \frac{x^{2}}{2}}{x^{3}}

O (x^{n}) \to 0 \Rightarrow \frac{\frac{x^{3}}{3!} - \frac{x^{3}}{3}}{x^{3}} = - \frac{1}{6}

Answered by mathmax by abdo last updated on 25/May/21

f (x) = \frac{e^{x} cosx - x - 1}{x^{3}} we have e^{x} \sim 1 + x + \frac{x^{2}}{2} and cosx \sim 1 - \frac{x^{2}}{2} \Rightarrow

e^{x} cosx \sim (1 + x + \frac{x^{2}}{2}) (1 - \frac{x^{2}}{2}) = 1 - \frac{x^{2}}{2} + x - \frac{x^{3}}{2} + \frac{x^{2}}{2} - \frac{x^{4}}{4}

= 1 + x - \frac{x^{3}}{2} - \frac{x^{4}}{4} \Rightarrow f (x) \sim \frac{1 + x - \frac{x^{3}}{2} - \frac{x^{4}}{4} - x - 1}{x^{3}} = - \frac{1}{2} - \frac{x}{4} \Rightarrow

\lim_{x \to 0} f (x) = - \frac{1}{2}