dstefan
03-22-2008, 01:06 PM
Chapter 1, Problem 4:
Use l'Hopital's rule to show that the following two Taylor approximations hold when x is close to 0:
\sqrt{1+x} \approx 1 ~+~ \frac{x}{2}
e^{x} \approx 1 ~+~ x ~+~ \frac{x^2}{2}
In other words, show that the following limits exist and are constant:
\lim_{x \to 0} \frac{\sqrt{1+x} ~-~ \left( 1 + \frac{x}{2} \right) }{x^2} and \lim_{x \to 0} \frac{e^{x} ~-~ \left( 1 + x + \frac{x^2}{2} \right)}{x^3}
Solution:
The numerator and denominator of each limit must be differentiated until a finite limit can be computed. Then l'Hopital's rule is applied sequentially to obtain the value of the initial limit:
\lim_{x \to 0} \frac{\sqrt{1+x} ~-~ \left( 1 + \frac{x}{2} \right) }{x^2} ~=~ \lim_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}} ~-~ \frac{1}{2}}{2x} ~=~ \lim_{x \to 0} \frac{-\frac{1}{4(1+x)^{3/2}}}{2} ~=~ -\frac{1}{8}
We conclude that \sqrt{1+x} ~=~ 1 ~+~ \frac{x}{2} ~+~ O(x^2)
\lim_{x \to 0} \frac{e^{x} ~-~ \left( 1 + x + \frac{x^2}{2} \right)}{x^3} ~=~ \lim_{x \to 0} \frac{e^{x} ~-~ (1 + x )}{3x^2} ~=~ \lim_{x \to 0} \frac{e^{x} ~-~ 1}{6x} ~=~ \lim_{x \to 0} \frac{e^{x}}{6} ~=~ \frac{1}{6}
We conclude that e^{x} ~-~ \left( 1 + x + \frac{x^2}{2} \right) ~=~ 1 ~+~ x ~+~ \frac{x^2}{2} ~+~ O(x^3)
Use l'Hopital's rule to show that the following two Taylor approximations hold when x is close to 0:
\sqrt{1+x} \approx 1 ~+~ \frac{x}{2}
e^{x} \approx 1 ~+~ x ~+~ \frac{x^2}{2}
In other words, show that the following limits exist and are constant:
\lim_{x \to 0} \frac{\sqrt{1+x} ~-~ \left( 1 + \frac{x}{2} \right) }{x^2} and \lim_{x \to 0} \frac{e^{x} ~-~ \left( 1 + x + \frac{x^2}{2} \right)}{x^3}
Solution:
The numerator and denominator of each limit must be differentiated until a finite limit can be computed. Then l'Hopital's rule is applied sequentially to obtain the value of the initial limit:
\lim_{x \to 0} \frac{\sqrt{1+x} ~-~ \left( 1 + \frac{x}{2} \right) }{x^2} ~=~ \lim_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}} ~-~ \frac{1}{2}}{2x} ~=~ \lim_{x \to 0} \frac{-\frac{1}{4(1+x)^{3/2}}}{2} ~=~ -\frac{1}{8}
We conclude that \sqrt{1+x} ~=~ 1 ~+~ \frac{x}{2} ~+~ O(x^2)
\lim_{x \to 0} \frac{e^{x} ~-~ \left( 1 + x + \frac{x^2}{2} \right)}{x^3} ~=~ \lim_{x \to 0} \frac{e^{x} ~-~ (1 + x )}{3x^2} ~=~ \lim_{x \to 0} \frac{e^{x} ~-~ 1}{6x} ~=~ \lim_{x \to 0} \frac{e^{x}}{6} ~=~ \frac{1}{6}
We conclude that e^{x} ~-~ \left( 1 + x + \frac{x^2}{2} \right) ~=~ 1 ~+~ x ~+~ \frac{x^2}{2} ~+~ O(x^3)