Failed Calculus Quiz: Did I Get the First Question Right?

[macbowes]

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

 

[arildno]

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!

 

[spamiam]

I quite like LaTeX! You mean:
<br /> \lim_{x \to -3} \frac{\sqrt{x^2}}{x}<br />?

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

 

[Ultimsolution]

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 )

 

[HallsofIvy]

\frac{\sqrt{x^2}}{x}= \frac{|x|}{x}[/tex<br /> If x&lt; 0, that is -1. If x&gt; 0, that is 1.<br /> <br /> Of course, if you are approaching -3, the only numbers you need to consider are negative numbers, so that is exactly the same as <br /> \lim_{x\to -3} -1<br /> <br /> (The problem would be more interesting if the limit were at x= 0.)

 

[Ultimsolution]

\frac{\sqrt{x^2}}{x}= \frac{|x|}{x}[/tex<br /> If x&lt; 0, that is -1. If x&gt; 0, that is 1.<br /> <br /> Of course, if you are approaching -3, the only numbers you need to consider are negative numbers, so that is exactly the same as <br /> \lim_{x\to -3} -1<br /> <br /> (The problem would be more interesting if the limit were at x= 0.)

<br /> <br /> <br /> <b><i>Of course that limit would not exist ...right?</i></b>

 

[gb7nash]

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...

 

[Ultimsolution]

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 doesn't exist...It doesnt... right?

 

[gb7nash]

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 doesn't 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.

 

[macbowes]

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:

<br /> lim_{x\to2}{{x^2-4}\over{x^3-8}}<br />

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.

 

[jhae2.718]

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.

 

[macbowes]

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:

<br /> \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