# Re: Using algebra when solving limits

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Posted by Joel on October 15, 2002 at 21:59:01:

In Reply to: Re: Using algebra when solving limits posted by mfrizza on October 15, 2002 at 20:03:16:

: Thanks for the explanation. I know this is an easy concept, but I'm still struggling to understand why the x becomes x^2 when you put it inside the sqrt. Why doesn't it just stay as x?

I don't understand what you don't understand. Are you quesioning HOW I made it x^2? x = sqrt(x^2)

Or are you asking WHY I wanted to do that? The WHY the the same as the reason why you're doing ALL of this manipulation: to end up with an expression that can be quanatified as x increases to infinity. If you have any terms of the form "infinity divided by infinity" you can't assign any value to them, so you have to "get rid" of all such terms by transforming them into something that you can evaluate. On the other hand, something in the form "integer divided by infinity" is OK, because that can be replaced by zero. So when I put the x inside the square root I was able to cancel terms and end up with just sqrt(9 + 1/x). As x approaches infinity the 1/x approaches zero, so the LIMITING VALUE of the sqrt term is just 3. What does that mean? It doesn't mean it will ever be exactly EQUAL to 3. But it does mean that I can make it get as close to 3 as I want (without quite getting all the way there), just by making x big enough.

: : : Help! I'm missing the point here...

: : : half way through this limits -> inf. problem and I'm stuck on a basic algebra issue...

: : : lim x-> inf.

: : : x / sqrt(9x^2+ x) + 3x

: : : next step...divide both sides by largest x in denominator...okay, let's use x

: : : x/x / sqrt(9x^2 + x)/x + 3x/x

: : : now we have 1 / sqrt(9x^2 + x)/x + 3...

: : : I don't understand the mechanics of sqrt(9x^2 + x)/x ....how does that work!

: : Assuming that 3x term is in the denominator, you should show it as x / (sqrt(9x^2+ x) + 3x).

: : The next step is to put that x inside the sqrt

: : = 1 / (sqrt((9x^2 + x)/x^2) + 3)
: : = 1 / (sqrt((9 + 1/x) + 3)
: : as x->infinity (1/x)->0 so you're left with
: : 1 / (sqrt(9) + 3) = 1/6

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