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# Evaluate the limit, if it exists. $\displaystyle \lim_{u \to 2}\frac{\sqrt{4u + 1}-3}{u - 2}$

## $\frac{2}{3}$

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to evaluate the limits of the square it. Of four. You plus one -3 Over U -2. As you approach us to note that by direct substitution we get The Square of four times 2 Plus 1 -3 Over 2 -2. This is equal to Scared of 9 -3/0. That's just 0/0. Now this value is indeterminate. That means we need to rewrite our function and so we have the limit as you approach is to of square the four U plus one minus three Over U -2. This time you will multiply it by the conjugate of the numerator to rationalize it. So we have times square it of For you plus one plus 3 over the same expression square at the four U plus one plus three. And so from here we have limit as you approach is to of the square of the square it Of for you plus one minus the square of three over u minus two Times the square at the four. You plus one plus three. Now simplifying, we have limit as you approach us to of four u plus 1 -9 over u minus two Times the square of the for you plus one plus three. And simplifying further we have a limit as you approach us to of four U -8 over u minus two Times Square. Therefore you plus one plus three. Now factory four in the numerator we have The limit as you approach is two of 4 times U -2 over U -2 times the square. Therefore you plus one Plus three. And here we can get rid of four of U -2. And so we have limit as you approach is to ah four over the square at the four U plus one plus three evaluating this. Now we have four over the square, the four times two plus one Plus three. That's the same as for over The squares of nine plus 3 Or for over six, which reduces to To over three. And so this is the value of the limit.

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