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.

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.

can anyone explain me on what basis did sal prove that the function does not exist by showing that wierd graph

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.

limx->infinity (sin1 ^n +cos 1 ^n )^n how to solve this sir?

You don't have an x anywhere in your function. It's just a constant.

At 2:07, the graph appears to be of a relation than a function? (Vertical Line Test)

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)

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.

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.

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Limits at infinity of quotients with trig (limit undefined)

Name: Limits at infinity of quotients with trig (limit undefined)
Uploaded: 2016-08-14T23:08:59Z
Description: 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.

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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.

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Kevin Vernon
Posted 8 years ago. Direct link to Kevin Vernon's post “So can you assume if the ...”
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.
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(6 votes)
Answer
- Vu
  Posted 8 years ago. Direct link to Vu's post “One should not assume. Es...”
  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.
  Comment on Vu's post “One should not assume. Es...”
  (12 votes)
Smith
Posted 7 years ago. Direct link to Smith's post “can anyone explain me on ...”
can anyone explain me on what basis did sal prove that the function does not exist by showing that wierd graph
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(6 votes)
Answer
- Kyle Gatesman
  Posted 7 years ago. Direct link to Kyle Gatesman's post “Since |sin(x)| <= 1, we c...”
  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.
  Button navigates to signup page
  (2 votes)
pratyushxman2
Posted 7 years ago. Direct link to pratyushxman2's post “limx->infinity (sin1 ^n +...”
limx->infinity (sin1 ^n +cos 1 ^n )^n how to solve this sir?
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(2 votes)
Answer
- kubleeka
  Posted 7 years ago. Direct link to kubleeka's post “You don't have an x anywh...”
  You don't have an x anywhere in your function. It's just a constant.
  Button navigates to signup page
  (8 votes)
Raivat Shah
Posted 8 years ago. Direct link to Raivat Shah's post “At 2:07, the graph appear...”
At
2:07
, the graph appears to be of a relation than a function? (Vertical Line Test)
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Juan Eduardo Ynsil Alfaro
  Posted 7 years ago. Direct link to Juan Eduardo Ynsil Alfaro's post “No way, sin(x) is a funct...”
  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)
  Button navigates to signup page
  (2 votes)
Kyle
Posted 6 years ago. Direct link to Kyle's post “Can i use squeeze theorem...”
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.
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(2 votes)
Answer
Vivek Rana
Posted 6 years ago. Direct link to Vivek Rana's post “At 1:22 sal says that eve...”
At
1:22
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.
Button navigates to signup pageComment on Vivek Rana's post “At 1:22 sal says that eve...”
(2 votes)
Answer
Jake
Posted 8 years ago. Direct link to Jake's post “Wouldnt it sometimes be d...”
Wouldnt it sometimes be defined, say when sin(x) = 1 coz itd be infinity?
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(2 votes)
Answer
- Neha Kumar
  Posted 7 years ago. Direct link to Neha Kumar's post “Yes, what you are saying ...”
  Yes, what you are saying is correct but infinity itself is undefined right?
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  (1 vote)
Karina
Posted 4 years ago. Direct link to Karina's post “For this question, you so...”
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
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(2 votes)
Answer
Zakoriya Ahmed
Posted 7 years ago. Direct link to Zakoriya Ahmed's post “So this does apply to cos...”
So this does apply to cosine if the function is (x^2+1)/cos(x) if I'm correct?
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(2 votes)
Answer
- kubleeka
  Posted 7 years ago. Direct link to kubleeka's post “Yes, it does. This become...”
  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.
  Button navigates to signup page
  (1 vote)
ＫＥＶＩＮ
Posted 2 years ago. Direct link to ＫＥＶＩＮ's post “I can kind of follow the ...”
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.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- kubleeka
  Posted 2 years ago. Direct link to kubleeka's post “Yes, the x's are always t...”
  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.
  Comment on kubleeka's post “Yes, the x's are always t...”
  (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.

Differential Calculus

Course: Differential Calculus > Unit 1

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Video transcript