# Finding Limit

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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arildno
Homework Helper
Gold Member
Dearly Missed
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!

2nafish117
I quite like LaTeX! You mean:
$$\lim_{x \to -3} \frac{\sqrt{x^2}}{x}$$?

There's another way to write $$\sqrt{x^2}$$ that might be helpful.

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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HallsofIvy
Homework Helper
$$\frac{\sqrt{x^2}}{x}= \frac{|x|}{x}[/tex If x< 0, that is -1. If x> 0, that is 1. Of course, if you are approaching -3, the only numbers you need to consider are negative numbers, so that is exactly the same as [tex]\lim_{x\to -3} -1$$

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

$$\frac{\sqrt{x^2}}{x}= \frac{|x|}{x}[/tex If x< 0, that is -1. If x> 0, that is 1. Of course, if you are approaching -3, the only numbers you need to consider are negative numbers, so that is exactly the same as [tex]\lim_{x\to -3} -1$$

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

Of course that limit would not exist ...right?

gb7nash
Homework Helper
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...

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?

gb7nash
Homework Helper
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.

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:

$$lim_{x\to2}{{x^2-4}\over{x^3-8}}$$

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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jhae2.718
Gold Member
What's a common term between $$x^2-4$$ and $$x^3-8$$ 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 $$x^2-4$$ that removes the difficulties caused as $$x \to 2$$ and see if you can factor it out of the denominator.

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What's a common term between $$x^2-4$$ and $$x^3-8$$ 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 $$x^2-4$$ that removes the difficulties caused as $$x \to 2$$ 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:

$$\lim_{x\to2}{{(x-2)(x+2)}\over{(x-2)(x^2+2x+4)}}$$

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

Got it, nvm :D

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