Finding the Limit of a Function: Sin(x)/x as x->0

[negation]

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]

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]

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