Proving Limit of (sin 5x)/x: Squeeze Theorem

[ktpr2]

$$f(x) = -50x^2+5$$
$$g(x) = (sin 5x)/x$$
$$h(x) = x^2+5$$

I'm trying to find the limit of g(x) as x --> 0

I know that f(x) and h(x) are less than and greater than, respectively, than g(x) but I am unsure how to prove that w/o abusing the concept of infinity. How would I prove this so that i can show that

$$f(x)\leqq g(x) \leqq h(x)$$

and use the squeeze theorem to show that the limit as x --> 0 for g(x) = 5 because it's also 5 for f(x) and h(x)? Alternatively, I'm sure, is there a better way to go about this?

 

[Data]

$$\lim_{x \rightarrow 0} \frac{\sin (5x)}{x} = \left(\frac{d}{dx} \sin (5x)\right) \biggr |_{x =0} = 5\cos{0} = 5.$$

Of course, you have to know a bunch of things to be able to use that to start with. Another easy way, if you already know

$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$$

then just multiply top and bottom of your limit by $5$ and note that $x\rightarrow 0 \Longleftrightarrow 5x \rightarrow 0$.

There's a thread on a similar limit in the math section that might give you some ideas (there's a geometrically motivated proof there too, using the squeeze theorem, if you want to do it that way):

https://www.physicsforums.com/showthread.php?t=70358

 

[thepatientmental]

The Squeeze Theorem states that if f(x) ≤ g(x) ≤ h(x) for all values of x near a (except possibly at a), and if lim f(x) = lim h(x) = L, then lim g(x) = L as x → a.

In this case, we can see that f(x) = -50x^2+5 is always less than g(x) = (sin 5x)/x for all values of x near 0 (except possibly at 0), since -50x^2+5 is a quadratic function that approaches 0 as x approaches 0, while (sin 5x)/x is a trigonometric function that oscillates between -1 and 1 as x approaches 0. Similarly, h(x) = x^2+5 is always greater than g(x) for all values of x near 0 (except possibly at 0), since x^2+5 is a quadratic function that approaches 5 as x approaches 0, while (sin 5x)/x oscillates between -1 and 1.

Therefore, we can conclude that -50x^2+5 ≤ (sin 5x)/x ≤ x^2+5 for all values of x near 0 (except possibly at 0). And since both -50x^2+5 and x^2+5 approach 5 as x approaches 0, we can use the Squeeze Theorem to show that the limit of g(x) as x approaches 0 is also equal to 5. This means that the limit of g(x) as x approaches 0 is indeed 5, and we have successfully proven it using the Squeeze Theorem.

In summary, the Squeeze Theorem is a powerful tool that allows us to find the limit of a function by "squeezing" it between two other functions whose limits we already know. In this case, we used the Squeeze Theorem to show that the limit of g(x) as x approaches 0 is 5 by showing that it is always between -50x^2+5 and x^2+5, both of which have a limit of 5 as x approaches 0. This is a valid and rigorous way to prove the limit without "abusing" the concept of infinity.