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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.

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• 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 , 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)
• 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.
• Wouldnt it sometimes be defined, say when sin(x) = 1 coz itd be infinity?
• 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
• So this does apply to cosine if the function is (x^2+1)/cos(x) if I'm correct?
• 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.