• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Squeeze theorem

  • Thread starter negation
  • Start date
Suppose I want to find the limit of a function g(x) as lim x-> 0

f(x) =< g(x) =<h(x)

My problem is selecting an upper and lower bound.

Suppose I have to find lim x->0 sin(x)/x

I know that sin(x) have a lower bound of -1 and upper bound of 1.

How should I further take this to solve the problem?
 

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
10,705
1,710
Suppose I want to find the limit of a function g(x) as lim x-> 0

f(x) =< g(x) =<h(x)

My problem is selecting an upper and lower bound.

Suppose I have to find lim x->0 sin(x)/x

I know that sin(x) have a lower bound of -1 and upper bound of 1.

How should I further take this to solve the problem?
Using your crude bounds, you can conclude that the limit of sin(x)/x lies between -∞ and +∞, if the limit exists at all. I think you will agree that this information is not very useful. Actually, you can improve it to lying between 0 and +∞, but that is not a whole lot better.

The limit is 100% standard, and appears in every calculus textbook and in many on-line calculus notes/tutorials.
 

pasmith

Homework Helper
1,667
369
Suppose I want to find the limit of a function g(x) as lim x-> 0

f(x) =< g(x) =<h(x)

My problem is selecting an upper and lower bound.

Suppose I have to find lim x->0 sin(x)/x

I know that sin(x) have a lower bound of -1 and upper bound of 1.

How should I further take this to solve the problem?
In practise one requires that [itex]\lim_{x \to 0} f(x) = \lim_{x \to 0} h(x)[/itex] for the squeeze therorem to be any use. However this problem is best solved by observing that
[tex]
\lim_{x \to 0} \frac{\sin(x)}{x}
[/tex]
is, if it exists, equal by definition to [itex]\sin'(0)[/itex]. If you already know by other methods (such as by formally defining [itex]\sin x[/itex] to be the solution of [itex]f'' + f = 0[/itex] which satisfies [itex]f(0) = 0[/itex] and [itex]f'(0) = 1[/itex]) that [itex]\sin'(0) = 1[/itex] then you're done, but otherwise you will have to proceed by looking at a suitable right-angled triangle and claiming that [itex]\sin x \approx x[/itex] for [itex]|x|[/itex] sufficiently small.
 

Want to reply to this thread?

"Squeeze theorem" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top