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This is problem number 64 of the stewart calculus 8th edition, section 2 .3.
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Evaluate the limit as x approaches 2 of the function square root of the quantity 6 minus x minus 2 divided by the square root of the quantity 3 minus x minus 1.
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We will first attempt to multiply by the conjugate of the numerator.
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Square root of 6 minus x plus 2 to the top and to the bottom.
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Let's see what this gives us.
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This gives us quantity 6 minus x minus 4 divided by the quantity square root of negative 3 minus x minus 1 times square root of 6 minus x plus 2.
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And then the top reduces to 6 minus 4, which is 2 minus x.
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So we have not yet made enough progress, as we are still not able to directly substitute 2 into this problem.
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So our approach will be to multiply by the conjugate of the bottom term here that we began with, this denominator, and see if that helps.
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The conjugate would be 3 minus x, and that quant...