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What is the lim as x approaches 0 of 3 times 1 - cosx divided by x?
First, you can take the constant factor 3 out, to obtain 3 times the limit of (1 - cos x) / x. Since this is of the form 0/0, you can use L'Hôpital's rule, which states that in such cases, you can take the derivative of both numerator and denominator. This results in the limit (as x approaches 0) of sin x / x, that is, 1 / 1 = 1. So, the final result is 3 times the limit of 1 = 3.
What is the limit as t approaches 0 of 2t tan t?
That would be 0. (But that was too easy. Did you mean something else?)
Does limit always tend to zero only?
No, limit can tend to any finite number including 0. It is also possible that the limit does not tend to any finite value or approaches infinity. Example: The limit of x^2+5 tend to 6 as x approaches -1.
Find the limit of lim sin 4x sin 6x x 0?
Unfortunately, the browser used by Answers.com for posting questions is incapable of accepting mathematical symbols. This means that we cannot see the mathematically critical parts of the question. We are, therefore unable to determine what exactly the question is about and so cannot give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "divided by", "equals" etc. As it appears, you seem to be seeking the limit of sin(4x)*sin(6x) as x tends to 0. Both components of the product tend to 0 as x tens to 0 and so the limit is 0. Bit I suspect that is not the limit that you are looking for.
2 sin squared x plus sin x minus 1 is equal to 0 solve for x?
2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210°, 330°or sin(x) - 1 = 0sin(x) = 1x = 90°
