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I understand almost everything until I get to the third line. If one multiplies the top and bottom by 1/x, where did the x^2 come from? Can someone explain.enter image description here

If you can't explain it simply, you don't understand it well enough. Albert Einstein

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    $\begingroup$ Irrelevant to the question, but that Einstein Quote is falsely attributed to him. $\endgroup$ – UserX Oct 12 '14 at 16:40
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I assume that you mean the $x^2$ denominators inside the square root. To "put the $x$ into" the square root, you must square it. For example:

$$\frac13\sqrt5=\sqrt{\frac59}$$

This is why the $x$ is squared.

By the way. Although your question has to be with a limit, it is not a limit question. This is algebra.

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  • $\begingroup$ Can you show me another way. I still do not understand. $\endgroup$ – user137452 Oct 12 '14 at 16:41
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Because he puts the (1/x) in the root, so 1/x=root(1/x^2). See it?

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  • $\begingroup$ What law of real number is this? $\endgroup$ – user137452 Oct 12 '14 at 16:44
  • $\begingroup$ If x is assumed positive, and you call y=root(1/x)=1/root(x), by definition you have: y^2=x $\endgroup$ – mvggz Oct 12 '14 at 16:48
  • $\begingroup$ @user137452 It follows from the rules $(a^b)^c = a^{bc}$ and $x^a y^a = (xy)^a$. You go from $x^{-1} x^{1/2}$ to $(x^{-2})^{1/2} x^{1/2}$ with the first rule. Then you go to $(x^{-1})^{1/2}$ with the second rule and simplification. $\endgroup$ – Ian Oct 12 '14 at 16:54
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$$\frac{\sqrt{81x^2+x}}x=\frac{\sqrt{81x^2+x}}{\sqrt{x^2}}=\sqrt{\frac{81x^2+x}{x^2}}=\sqrt{81\frac{x^2}{x^2}+\frac x{x^2}}=\sqrt{81+\frac1x}$$

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