If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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?

  • blobby green style avatar for user Kevin Vernon
    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)
    Default Khan Academy avatar avatar for user
    • duskpin sapling style avatar for user Vu
      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)
  • piceratops ultimate style avatar for user Smith
    can anyone explain me on what basis did sal prove that the function does not exist by showing that wierd graph
    (6 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Kyle Gatesman
      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)
  • blobby green style avatar for user pratyushxman2
    limx->infinity (sin1 ^n +cos 1 ^n )^n how to solve this sir?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Raivat Shah
    At , the graph appears to be of a relation than a function? (Vertical Line Test)
    (2 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user Juan Eduardo Ynsil Alfaro
      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)
  • leafers seed style avatar for user Kyle
    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)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Jake
    Wouldnt it sometimes be defined, say when sin(x) = 1 coz itd be infinity?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • female robot ada style avatar for user Karina
    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)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user Zakoriya Ahmed
    So this does apply to cosine if the function is (x^2+1)/cos(x) if I'm correct?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      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)
  • male robot hal style avatar for user KEVIN
    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)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      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)
  • primosaur sapling style avatar for user Vivek Rana
    At sal 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.
    (1 vote)
    Default Khan Academy avatar avatar for user

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.