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Mathematics 23 Online
OpenStudy (anonymous):

Evaluate the limit

OpenStudy (anonymous):

OpenStudy (anonymous):

Two options. Either multiply top and bottom by the conjugate of the bottom, or take advantage of difference of squares: \[a-b = (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})\]

OpenStudy (anonymous):

ok i think id rather take the conjugate

OpenStudy (anonymous):

Either works :) The difference of squares option isn't as obvious when you're not used to it with things that aren't actually like \(x^{2}-4\). But it can be done with the anything that is a-b form.

OpenStudy (anonymous):

the conjugate is \[1+\sqrt{1}\] right?

OpenStudy (anonymous):

Top and bottom by \(1+\sqrt{b}\)

OpenStudy (anonymous):

ok once we do that do i multiply them out?

OpenStudy (anonymous):

Multiply out the bottom one. Multiplying out the top one makes it hard to see what happens.

OpenStudy (anonymous):

ok so how does the bottom look like?

OpenStudy (anonymous):

1-b?

OpenStudy (anonymous):

then that cancels out with the top right ?

OpenStudy (anonymous):

\[\frac{ (1-b)(1+\sqrt{b}) }{ (1-\sqrt{b})(1+\sqrt{b}) } = \frac{ (1-b)(1+\sqrt{b}) }{ 1+\sqrt{b}-\sqrt{b}-b } = \frac{ (1-b)(1+\sqrt{b}) }{ 1-b }\]

OpenStudy (anonymous):

Yes, then it would cancel :)

OpenStudy (anonymous):

then 1+\[\sqrt{b}\]

OpenStudy (anonymous):

would be left over right?

OpenStudy (anonymous):

Yep. And then you can plug in your limit.

OpenStudy (anonymous):

so the answer would be 2

OpenStudy (anonymous):

Yes :3

OpenStudy (anonymous):

ah yes thank you

OpenStudy (anonymous):

You're welcome :)

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