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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 1

Lesson 20: Limits at infinity (horizontal asymptotes)- Infinite limits intro
- Limits at infinity of quotients (Part 1)
- Limits at infinity of quotients (Part 2)
- Limits at infinity of quotients
- Limits at infinity of quotients with square roots (odd power)
- Limits at infinity of quotients with square roots (even power)
- Limits at infinity of quotients with square roots
- Limits at infinity of quotients with trig
- Limits at infinity of quotients with trig (limit undefined)
- Limits at infinity of quotients with trig
- Limit at infinity of a difference of functions

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# Limits at infinity of quotients with trig (limit undefined)

Sal analyzes the limit of (x²+1)/sin(x) at infinity. It turns out this limit doesn't exist, as the function keeps oscillating between positive and negative infinity.

## Want to join the conversation?

- So can you assume if the trig function is the denominator and by itself and the numerator is a function of x but not a trig function then the limit does not exist.(6 votes)
- One should not assume. Especially math, we require proofs. While most have diverged limits, there is one infinite limit (technically doesn't exist but we know their end behaviors). Then it also depends on what x approaches to when we are asked to find the limit. While the limit at infinity may not exist, its limit at a specific x may exist.(12 votes)

- can anyone explain me on what basis did sal prove that the function does not exist by showing that wierd graph(6 votes)
- Since |sin(x)| <= 1, we can say the absolute value of the limit must be at least (x^2 + 1) / (1). This already goes to infinity, so the lower bound for the absolute value of the limit does not exist. Furthermore, sin(x) oscillates between positive and negative values, so the limit essentially oscillates between +infinity and -infinity. Therefore, we cannot describe its end behavior as exclusively +infinity or exclusively -infinity.(2 votes)

- limx->infinity (sin1 ^n +cos 1 ^n )^n how to solve this sir?(2 votes)
- You don't have an x anywhere in your function. It's just a constant.(6 votes)

- At2:07, the graph appears to be of a relation than a function? (Vertical Line Test)(2 votes)
- No way, sin(x) is a function (trigonometric function); x^2+1 is a function too. The ratio between two fuctions should be also a function. I've already made the vertical line test on that function and a I didn't find nothing. This is because every value of x takes only a value of y. Every x gives only a y=sin(x). Every x gives only a x^2+1. Every x gives only a x^2+1/sin(x)(2 votes)

- Can i use squeeze theorem to show that the left limit on the theorem is negative infinity and the right is infinity, and therefore by squeeze theorem not existent?

I'm trying to find a short concise answer that would suffice for explaining such on an exam with this question and no graph calculator.(2 votes) - At1:22sal says that every time when sin x becomes zero we are gonna have a vertical asymptote. Why does he say that please explain. And is there any video in Khan Academy particularly on asymptotes? If there is please suggest.(2 votes)
- Wouldnt it sometimes be defined, say when sin(x) = 1 coz itd be infinity?(2 votes)
- Yes, what you are saying is correct but infinity itself is undefined right?(1 vote)

- For this question, you solved it graphically. Is it possible to use the same approach as the video "Limits at infinity of quotients with trig" and set up a double inequality like

(x^2+1)/-1 < (x^2+1)/sinx < (x^2+1)/1(2 votes) - So this does apply to cosine if the function is (x^2+1)/cos(x) if I'm correct?(2 votes)
- Yes, it does. This becomes more obvious if we rewrite your function as (x²+1)/sin(π/2-x)= -(x²+1)/sin(x-π/2).

The tiny shift of π/2 won't matter in a limit at infinity, and if a limit is undefined, then so, of course, is its negative.(1 vote)

- I can kind of follow the logic Sal uses in the video to "see" his solution. My question in general though; is the x input in "sin(x)" the same as the x input in "(x^2+1)"? If it is not, why would we use two different x-values, and if it is, how would we know that that is the case? I've tried various inputs to see if I can match the (x,y) points as plotted at Desmos, but can't seem to make the connection(s). Thanks for any input/answers.(1 vote)
- Yes, the x's are always the same. That's why they're represented as the same variable. If we wanted to allow them to be different, we would have to represent them with two different variables, such as sin(x)/(y²+1).

In this case, you would be dealing with a multi-variable function, which is not covered until much later.(2 votes)

## Video transcript

- [Voiceover] Let's see
if we can figure out what the limit of x squared
plus one over sine of x is as x approaches infinity. So, let's just think about what's going on in the numerator and then
think about what's going on in the denominator. So, the numerator, we
have x squared plus one. So, as x gets larger and larger and larger as it approaches infinity,
well, we're just squaring it here so this
numerator's gonna get even, approach infinity even faster. So, this thing is going to go to infinity as x approaches infinity. Now what's happening to
the denominator here? Well, sine of x, we've seen this before. Sine of x and cosine of x are bounded, they oscillate. They oscillate between
negative one and one, so negative one is gonna
be less than or equal to sine of x which is going to
be less than or equal to one. So, this denominator's going to oscillate. So, what does that tell us? Well, we might be tempted
to say, well the numerator's unbound and goes to infinity
and the denominator's just oscillating between
these values here. So maybe the whole thing goes to infinity. But we have to be careful because one, the denominator's going between positive and negative values. So, the numerator's just going to get more and more and more positive being divided sometimes
by a positive value, sometimes by a negative value. So, we're gonna jump between
positive and negative. Positive and negative. And then you also have all
these crazy asymptotes here. Every time sine of x becomes zero, well then, you're gonna
have a vertical asymptote. This thing will not be defined. So you have all these vertical asymptotes. You're gonna oscillate
between positive and negative just larger and larger values. So, this limit does not exist. So, it does not exist. Does not exist. And we can see that graphically. We described it in words, just inspecting this expression, but we can see it
graphically if we actually looked at a graph of this, which I have right here. And you can see that as x goes
towards positive infinity, as x goes to positive infinity, we, depending on which x we
are, we're kind of going, we go, we get really large, then we had a vertical asymptote than
we jump back down and go really negative, vertical asymptote, up, down, up, down, up, down, the oscillations just
get more and more extreme and we keep having these
vertical asymptotes on a periodic basis. So it's very clear that
this limit does not exist.