WEBVTT
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Find the limit as π₯ approaches zero of three sin four π₯ over two cos π₯.
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In order to find a limit, our first instinct might be to use direct substitution.
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Letβs see how we could go about doing this in this question.
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Recall that, for all constants π and π, their functions π sin ππ₯ and π cos ππ₯ are continuous on the set of all real numbers.
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Therefore, both the functions three sin four π₯ in the numerator and two cos π₯ in the denominator of the limit weβre asked to find are continuous on the set of all real numbers.
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In particular, the functions three sin four π₯ and two cos π₯ are continuous at π₯ equals zero.
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Therefore, their quotient three sin four π₯ over two cos π₯ is continuous at π₯ equals zero, given that two cos of zero is not equal to zero.
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We have that cos of zero equals one.
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Therefore, two cos of zero equals two, which is not equal to zero.
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Therefore, the quotient three sin four π₯ over two cos π₯ is continuous at π₯ equals zero.
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Now recall that π of π₯ is continuous at π₯ equals π implies that the limit of π of π₯ as π₯ approaches π equals π evaluated at π.
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Therefore, we can use direct substitution to find the limit given to us in the question.
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Doing so, we obtain that the limit as π₯ approaches zero of three sin four π₯ over two cos π₯ is equal to three sin of four times zero over two cos of zero.
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Which simplifies to three sin of zero over two cos of zero.
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Using the fact that sin of zero equals zero and cos of zero equals one, this is equal to three times zero over two times one.
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Which simplifies to zero over two, which is just zero.
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So we have found that the limit as π₯ approaches zero of three sin four π₯ over two cos π₯ is equal to zero.