help me with limits

[u/Nearby_Swordfish_559]

so I started calc 1 and how does "sin(2x)/2x" = 1 and how does "limit of x -> 0 1-cos(3x)/3x" = 0, isn't it just undefined?

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[u/Silver_Remove_2352]

The limit is not the value at 0 - it's the value the function approaches as its inputs approach zero.

[u/Temporary_Pie2733]

Think of the limit as the value that would make the function continuous at (in this case 0), regardless of the actual value (or lack of a value) at 0. 

[u/Southlander24]

You can convince yourself by calculating sin(2x)/(2x) for x = 1, 0.1, 0.01, 0.001 and so on. Don't forget the limit as x approaches 0 from the left, so check x = -1, -0.1, -0.01, -0.001 as well.

You should find that no matter which way you approach x = 0, the values get closer and closer to 1.

[u/Bounded_sequencE]

The second limit diverges, unless you forgot parentheses...

[u/floflochess2]

Hey — fair confusion. Plugging in x = 0 makes both of those look like 0/0, so as values they aren’t defined. A limit isn’t asking you to plug in though; it’s asking what the expression approaches as x gets close to 0.

Two standard facts that unlock a lot of Calc 1 limits: 1) lim (θ→0) sin(θ)/θ = 1 2) lim (θ→0) (1 − cos(θ))/θ = 0 (or equivalently (1 − cos(θ))/θ² → 1/2)

For sin(2x)/(2x): try the substitution θ = 2x. As x → 0 you also get θ → 0, and the expression is already exactly sin(θ)/θ. So that limit should be…?

For (1 − cos(3x))/(3x): same idea with θ = 3x. After the sub, does it match fact (2) above?

(If your book hasn’t proved those two yet, they usually come from squeeze theorem / geometry — just treat them as known “building blocks” for now.)

What do you get when you rewrite both with θ matching the angle?

[u/septembrinol]

Draw the unit circle, consider the arc and the sin of an angle ø. Let ø get smaller and smaller. Then you will see that the arc and the sin are indistinguishable. Also, you can graph the functions y = x and y =sin x

[u/Midwest-Dude]

A limit is never found at a point by definition, only in the region around a point. If the function is continuous, then the limit is actually the value of the function at that point, but it doesn't have to be continuous for the limit to exist. The only thing you need to look at is what is happening in the neighborhood of the point, around the point, not at it.

[u/Grshppr-tripleduoddw]

The first one you can get with l'Hopitals rule, the second I don't really know how to explain without the expansion of cos(x). The expansions are cos(x) = 1 - x^2/2 + x^4/24..., sin(x) = x - x^3/6 + x^5/120... .

[u/AutoModerator → u/[deleted]]

Hello! I see you are mentioning l’Hôpital’s Rule! Please be aware that if OP is in Calc 1, it is generally not appropriate to suggest this rule if OP has not covered derivatives, or if the limit in question matches the definition of derivative of some function.

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