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# Limits at infinity of quotients with trig

Sal finds the limit of cosx/(x²-1) at infinity, by putting it between two limits of rational functions, 1/(x²-1) and -1/(x²-1).

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• I, assume, though he doesn't mention it in the video that we are applying the Pinching Theorem (Squeeze or Sandwich, if you prefer)...
• Is there an even mathy way (as suggested in the video by Sal) to prove this? Probably using principle mathematical induction or something?
• At , why is the inequality -1/(x²-1) <= cosx/(x²-1) <= 1/(x²-1) true for all x?
For example, if x = 0, then -1/((0)²-1) <= cos(0)/((0)²-1) <= 1/((0)²-1) becomes 1 <= -1 <= -1, which is false, right?
• You're absolutely right. It seems like a mistake on Sal's part. I think it should have been something like 0 ≤ | cos(x)/(x^2 -1) | ≤ | 1/(x^2 -1) |. Good catch.
• it is like squize theorem , right?
• wait, so irrespective of the sign of the numerator, as long as the value of the denominator becomes increasingly negative or increasingly positive, the limit as x --> infinity & x--> -infinity will equal to zero?
• Yes, you are correct. But to be clear, as long as the denominator becomes sufficiently LARGE as compared to a relatively small numerator (whether positive or negative), the limit as x->infinity will be 0.

Remember, a tiny numerator (negative or positive) divided by a HUGE denominator (negative or positive) will be very close to zero. That's why the limit as the number goes to infinity is zero.
• how does it become a negative?...when its to the power of an odd ?
(1 vote)
• there is a mistake: you can't say that if -1<cosx<1 then -1/x*2-1<cosx/x*2-1<1/x*2-1; because if x*2-1 is negative than you would have to invert the diseqution
• He's talking about their "limiting value" when x approaches infinity :)
(1 vote)
• At x equals +1 or -1, the function goes undefined. Shouldn't we consider that point as we approach Infinity ?
• Your point is correct. However, when we talk about taking the limit "as x approaches infinity," we should really be thinking of "the value of the function AT x = infinity." Also, limits are only concerned with the point we end at (infinity in this case), not where we "start approaching from." Anyway, when we think of limits "AT" infinity, we plug in x = infinity into the expression, remembering that p*x^a + q*x^b =
(1) p*x^a if a > b,
(2) q*x^b if b > a, or
(3) (p + q)*x^a if a = b.
The above conditions hold true at x = infinity and allow us to reduce the problem to a limit that is smaller in magnitude than 1/x^2, which abruptly flattens to 0 as x becomes large.