# Calc Problem

• Dec 24th 2008, 11:51 AM
calc_help123
Calc Problem
The function f is defined by

f(x)=
(sinx)/(x) , x ≠ 0

1, x=0

A) at what points does its graph cross the x-axis?
B) What is the relation between x and tanx at points x≠0 at which f'(x)=0
C)What is the behavior of f as the absolute value of x approaches infinite
• Dec 24th 2008, 12:22 PM
galactus
A: $\frac{sin(x)}{x}=0$

$x=C{\pi}$

B: $f'(x)=\frac{cos(x)}{x}-\frac{sin(x)}{x^{2}}$

$\frac{cos(x)}{x}=\frac{sin(x)}{x^{2}}$

$x^{2}cos(x)=xsin(x)$

$x=\frac{sin(x)}{cos(x)}=tan(x)$

C:

$\lim_{x\to {\infty}}\frac{sin|x|}{|x|}$

$\lim_{x\to {\infty}}\frac{sin(-x)}{-x}=\lim_{x\to {\infty}}\frac{-sin(x)}{-x}=\lim_{x\to {\infty}}\frac{sin(x)}{x}$

See the limit?.
• Dec 31st 2008, 05:40 PM
OnMyWayToBeAMathProffesor
sry, just to clarify, $
\lim_{x\to {\infty}}\frac{sin|x|}{|x|}
$
is the same as $
\lim_{x\to {\infty}}\frac{sinx}{x}
$
which is zero, correct?
• Dec 31st 2008, 06:27 PM
Jhevon
Quote:

Originally Posted by OnMyWayToBeAMathProffesor
sry, just to clarify, $
\lim_{x\to {\infty}}\frac{sin|x|}{|x|}
$
is the same as $
\lim_{x\to {\infty}}\frac{sinx}{x}
$
which is zero, correct?

yes. since |x| = x when x is positive. and the limit goes to zero. you can prove this using the squeeze theorem. one of many ways