How to find the limit of $$\lim_{x \to \infty} 3\cos(2x +1)+ [1/(2x +1)^3 ] - 1$$ when $x$ goes to infinity. When I am doing this question I couldn't able to find the limit of cos(2x +1). Should I put a constant for cos(2x +1) or can I calculate the limit without using a constant ?
How to calculate a limit when x goes to infinity without using l'hospital 's rule.
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limits
infinity
limits-without-lhopital
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0$\cos(2x+1)$ has no limit at $\infty$ since it's a periodic function – 2017-01-31
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0But if I have to calculate it can I calculate it using k(constant)? Whatever the limit the cos(2x +1) should be in between -1 & +1. Is that correct? – 2017-01-31
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0If it *had* a limit, it would be between $-1$ and $1$, but it hasn't. – 2017-01-31
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0Then isn't there an answer for the above question? – 2017-01-31
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0The answer is there's no limit. – 2017-01-31
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0But I have to calculate the limit of above function. Then how to calculate that thing? – 2017-01-31
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0We can't determine a limit that doesn't exist! Are you sure the formula you posted is exact? – 2017-01-31
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0Yes it is. I have to find the value of it. Don't you have any idea of evaluating that thing? – 2017-01-31
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0I only can tell it's impossible. – 2017-01-31
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0When I add k for the limit of cos(2x +1) . I got an answer like 3k -1. Is that correct?. Can you give me an answer assuming that there's a limit for cos (2x +1). – 2017-01-31
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0Yes, but why should $\cos(2x+1)$ tend to $k$ when $x$ tends to $\infty$? – 2017-01-31
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0Because I thought that cos(2x +1) should always have a value in between -1 & +1. So k should be a constant in between -1 & +1. Is that wrong? – 2017-01-31
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0I already answered that if it had a limit, it would be between $-1$ and $1$. The problem is that it hasn't one. Just like you can suppose God has a lot of properties if it exists. But does it? – 2017-01-31
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0I have no idea. But I have to find the answer for the above question. What should I write as the answer? – 2017-01-31
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0Could post a (link to a) scan of the full real question? – 2017-01-31
1 Answers
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The limit does not exist.
If we assume the limit exists, then $\lim\limits_{x \to \infty} \cos x$ must exist. However, the limit does not exist, as it moves between $-1$ or $1$.
So the answer is the limit does not exist.