Learn how to evaluate the limit at infinity of a trigonometric function

👉 We will explore how to evaluate the limit at infinity. When evaluating the limit at infinity or negative infinity we are interested to know where is the graph going right and left. This is also commonly explored as end behavior of the graph. Most of the examples we will look at will include rational functions. When looking at where the graph is going right or left we need to identify the horizontal asymptotes. We will also take a look at rational functions with roots where we will need to look at more closely at how the root behaves with the graph.

Organized Videos:
✅The Limit
✅Evaluate Limits of Complex Fractions
✅Evaluate Limits of Polynomials
✅Evaluate Limits of Rational Expressions
✅Evaluate Limits with Square Roots
✅Evaluate Limits with Trig
✅Limits of Piecewise Functions
✅Evaluate Limits with Transcendentals
✅Evaluate Limits Difference Quotient
✅Evaluate Limits from a Graph
✅Evaluate Limits of Absolute Value
✅Evaluate Limits of Square Root
✅Holes and Asymptotes of Rational Functions
✅Learn about Limits
✅Find the Value that makes the Function Continuous
✅Is the Functions Continuous or Not?
✅Evaluate Limits using a Table of Values
✅Evaluate Limits at Infinity

Connect with me:

[mateuszgawronski]

There is a mistake, you broke apart sin(2x)/x into (1/x) * (sin(2x)/x) and it should be just (1/x)*(sin(2x)) :)
1:03 in the video :)

[lukemilitzer]

your videos saved me in 10th grade, and now they are saving me in college

[b.blokzijl]

The answer is correct but the method is not. This problem has to be solved with the squeeze theorem,
-1 <= sin(2x) <= 1
-1/x <= sin(2x) / x <=1/x
For the limit x approaches infinity we have:
0 <= sin(2x) /x <= 0
Squeeze theorem gives us:
Limit as x approaches infinity of sin(2x) / x approaches 0.

[saranmin]

I honestly love your videos please keep doing what youre doing.. youve helped me through highschool soooo much

[Joffrerap]

the problem is that 0*DNE is not necessarly 0. only way to do this is squeeze theoreme like someone did in the comments. This is like the bad physician way of doing this.

[Jaunter]

Thank You Mr Brian. You have saved me yet again

[simon-ghpt]

Range of sinx function is always [-1, 1] so if we divide any number from -1to 1 by a number x which approaches infinity then sinx/x approaches 0, it's very easy I think.

[erinakii]

question: what level is this...as in in which highschool class is this taught? or do students learn this in college?

[kuroao]

The answer is corect, but there is wrong step. You should multiply the function by 2x/2x, then you have (sin2x)/2x * 2x/x=(sin2x)/2x * 2. Limit x to inf of (sin2x)/2x is 0, and limit x to inf of 2 is 2. So 0*2 is 0
Thanks

[kagisomarvinmolekwa]

I don't understand. What don't we use the identity: sin(x)/x = 1

sin(2x)/x * 2/2 = 2sin(2x)/2x

which equals 2

[pravinnitb]

Not long it short at 2:08 but nice explain

[b.blokzijl]

how can you multiply DNE by 0? I mean, how can you mutiply DNE by any realnumber, DNE is not a number .... I did graph the function and indeed with this limit the function approaches 0. But can someone explain to me why it is justified to multiply by DNE in this case? Isn't this proof simply incorrect?