Finding Limit

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Hey all, I just finished (aka, failed epically) a calculus quiz about 10 minutes ago. I'm in my school library right now and I'm just wondering if I got the first question of the quiz right.

I'm just going to say what the question was since latex is bloody impossible to use.

limit of x -> -3 when square root of x2 is over x
 
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  • #2
arildno
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Hey all, I just finished (aka, failed epically) a calculus quiz about 10 minutes ago. I'm in my school library right now and I'm just wondering if I got the first question of the quiz right.

I'm just going to say what the question was since latex is bloody impossible to use.

limit of x -> -3 when square root of x2 is over x
Tell me what you answered, and I'll tell you if it was right! :smile:
 
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I quite like LaTeX! You mean:
[tex]
\lim_{x \to -3} \frac{\sqrt{x^2}}{x}
[/tex]?

There's another way to write [tex] \sqrt{x^2} [/tex] that might be helpful.
 
  • #4
Hey, here is how i would approach the problem :

Lets start the the square root part , squareroot of (X^2) = absolute value of x ,

now as x approaches -3 , x is negative which means that absolute value of x = -x ( hopefully you see where i'm going with this.

Now what you've got here basically comes down to lim (as x approaches -3) of (-x/x) which is -1.

Another less elaborate and more straightforward way of doing this is by simply plugging in -3, you also get -1 as the answer (and this can be justified by the fact that the function squareroot of X^2 and the function 1/x are both continuous at -3 )
 
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  • #5
HallsofIvy
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[tex]\frac{\sqrt{x^2}}{x}= \frac{|x|}{x}[/tex
If x< 0, that is -1. If x> 0, that is 1.

Of course, if you are approaching -3, the only numbers you need to consider are negative numbers, so that is exactly the same as
[tex]\lim_{x\to -3} -1[/tex]

(The problem would be more interesting if the limit were at x= 0.)
 
  • #6
[tex]\frac{\sqrt{x^2}}{x}= \frac{|x|}{x}[/tex
If x< 0, that is -1. If x> 0, that is 1.

Of course, if you are approaching -3, the only numbers you need to consider are negative numbers, so that is exactly the same as
[tex]\lim_{x\to -3} -1[/tex]

(The problem would be more interesting if the limit were at x= 0.)

Of course that limit would not exist ...right?
 
  • #7
gb7nash
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Of course that limit would not exist ...right?
Why do you think that? Ignoring the signs, the numerator is the same as the denominator. When you approach -3, the numerator is positive and the denominator is negative. So...
 
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Why do you think that? Ignoring the signs, the numerator is the same as the denominator. When you approach -3, the numerator is positive and the denominator is negative. So...
I am not discussing here the limit when x approaches -3...it's been estblished many posts ago that it is -1... When HalsoIvy said that it would be a lot more interesting to evaluate that limit when x---> 0, he/she awakened my curiosity and I evaluated it and found thatit doesnt exist...It doesnt... right?
 
  • #9
gb7nash
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I am not discussing here the limit when x approaches -3...it's been estblished many posts ago that it is -1... When HalsoIvy said that it would be a lot more interesting to evaluate that limit when x---> 0, he/she awakened my curiosity and I evaluated it and found thatit doesnt exist...It doesnt... right?
Ah, my mistake. You would be right, if it approaches 0. If you approach from the left, you get -1, if you approach from the right, you get +1.
 
  • #10
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Hey, here is how i would approach the problem :

Lets start the the square root part , squareroot of (X^2) = absolute value of x ,

now as x approaches -3 , x is negative which means that absolute value of x = -x ( hopefully you see where i'm going with this.

Now what you've got here basically comes down to lim (as x approaches -3) of (-x/x) which is -1.

Another less elaborate and more straightforward way of doing this is by simply plugging in -3, you also get -1 as the answer (and this can be justified by the fact that the function squareroot of X^2 and the function 1/x are both continuous at -3 )

This is basically what I answered. I ended up getting -1, but I didn't even know what that -1 meant, and I had no idea if I was even close to right. I think my trouble is that I don't understand limits or what they are, so it makes answering questions about them difficult.

EDIT: Okay, so I know I got that question right. The next question was:

[tex]
lim_{x\to2}{{x^2-4}\over{x^3-8}}
[/tex]

I tried to complete the square on the top, but then I couldn't figure out how to factor the bottom. Yes, I hate factoring. I got that far, got really frustrated, then just packed up my stuff and handed my prof my quiz.
 
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  • #11
jhae2.718
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What's a common term between [tex]x^2-4[/tex] and [tex]x^3-8[/tex] that you can factor out? It may help to review how a difference of two cubes is factored, or you could just do something like synthetic division. Hint: look for the factor of [tex]x^2-4[/tex] that removes the difficulties caused as [tex]x \to 2[/tex] and see if you can factor it out of the denominator.
 
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  • #12
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What's a common term between [tex]x^2-4[/tex] and [tex]x^3-8[/tex] that you can factor out? It may help to review how a difference of two cubes is factored, or you could just do something like synthetic division. Hint: look for the factor of [tex]x^2-4[/tex] that removes the difficulties caused as [tex]x \to 2[/tex] and see if you can factor it out of the denominator.
Well, if I write out the question factored, here's where I got to:

[tex]
\lim_{x\to2}{{(x-2)(x+2)}\over{(x-2)(x^2+2x+4)}}[/tex]

I guess I could cancel the (x - 2)'s, but then what?

Got it, nvm :D
 
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