Question

find the one sided limit of the function f(x)=|sin x|/x

Answer 1

i know that sinx/x=1 , but how will you apply this to the question?

Answer 2

First of all, what is \(x\) approaching?

Second, which side? Or are you computing both one-sided limits?

Answer 3

x approaches 0. computing both sides

Answer 4

Recall the absolute value definition:
\[|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}\]
This means
\[|\sin x|=\begin{cases}\sin x&\text{for }x\ge0\\-\sin x&\text{for }x<0\end{cases}\]
For the left-sided limit, you would use the part of \(|\sin x|\) defined for \(x<0\). For the right, you would use the part defined for \(x\ge0\).

Answer 5

so substituting x=0, both the left and the right will be zero?

Answer 6

Or you have to use L'hospital rule

Answer 7

Not quite, you have
\[\lim_{x\to0^+}\frac{|\sin x|}{x}=\lim_{x\to0^+}\frac{\sin x}{x}=1\]
and
\[\lim_{x\to0^-}\frac{|\sin x|}{x}=\lim_{x\to0^-}\frac{-\sin x}{x}=-\lim_{x\to0^-}\frac{\sin x}{x}=-1\]

Answer 8

so in case x approaches 1, how do i go about it?

Answer 9

\(\dfrac{\sin x}{x}\) is continuous everywhere except where \(x=0\), so you can use the direct-substitution rule,
\[\lim_{x\to c}f(x)=f(c)\]
if \(f(x)\) is continuous at \(c\).

Answer 10

So if you had
\[\lim_{x\to1}\frac{|\sin x|}{x}\]
you would substitute, giving
\[\frac{|\sin 1|}{1}=\sin 1\]

Answer 11

if x approaches any other number apart from 0 you can use the direct substitution. Is that right?