It can be shown that the number e satisfies the inequality ( 1 + 1 n ) n < e < ( 1 + 1 n ) n + 1 n ≥ 1 Illustrate this condition by graphing y 1 = ( 1 + 1 / n ) n y 2 = 2.718 281 828 ≈ e y 3 = ( 1 + 1 / n ) n + 1 in the same viewing window, for 1 ≤ n ≤ 20.
It can be shown that the number e satisfies the inequality ( 1 + 1 n ) n < e < ( 1 + 1 n ) n + 1 n ≥ 1 Illustrate this condition by graphing y 1 = ( 1 + 1 / n ) n y 2 = 2.718 281 828 ≈ e y 3 = ( 1 + 1 / n ) n + 1 in the same viewing window, for 1 ≤ n ≤ 20.
Solution Summary: The author illustrates how the number e satisfies the inequality (1+1n) n+1 by plotting the graph of the given functions.
A function is defined on the interval (-π/2,π/2) by this multipart rule:
if -π/2 < x < 0
f(x) =
a
if x=0
31-tan x
+31-cot x
if 0 < x < π/2
Here, a and b are constants. Find a and b so that the function f(x) is continuous at x=0.
a=
b= 3
Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
f(x) = (x + 4x4) 5,
a = -1
lim f(x)
X--1
=
lim
x+4x
X--1
lim
X-1
4
x+4x
5
))"
5
))
by the power law
by the sum law
lim (x) + lim
X--1
4
4x
X-1
-(0,00+(
Find f(-1).
f(-1)=243
lim (x) +
-1 +4
35
4 ([
)
lim (x4)
5
x-1
Thus, by the definition of continuity, f is continuous at a = -1.
by the multiple constant law
by the direct substitution property
4 Use Cramer's rule to solve for x and t in the Lorentz-Einstein equations of special relativity:x^(')=\gamma (x-vt)t^(')=\gamma (t-v(x)/(c^(2)))where \gamma ^(2)(1-(v^(2))/(c^(2)))=1.
Chapter 3 Solutions
Calculus for Business, Economics, Life Sciences, and Social Sciences (13th Edition)
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