Finding the limit horizontal asymptotes

[ScienceJen]

Hi everyone,

I'm finding calculus difficult...can someone please riddle me this? How does x^2 become x^2 in the denominator from the numerator and denominator being multiplied by 1/x?
I understand everything else, I think. Does x automatically become squared under the sqrt? That doesn't make sense.

Thank-you. Here's a screenshot. I'll talking about question #21.

 

[Deleted member 4993]

Hi everyone,

I'm finding calculus difficult...can someone please riddle me this? How does x^2 become x^2 in the denominator from the numerator and denominator being multiplied by 1/x?
I understand everything else, I think. Does x automatically become squared under the sqrt? That doesn't make sense.

Thank-you. Here's a screenshot. I'll talking about question #21.

View attachment 31800

​

To answer your general question:

Does x automatically become squared under the sqrt?

​

You do know that 3² = 9 and √9 = 3 ??

3 = √9 = √(3²) = 3

similarly

x = \(\displaystyle \sqrt{x^2} \)..........................correct expression would be →

|x| = \(\displaystyle \sqrt{x^2} \)

​

So to quote you - yes ... x (if it is a positive number) automatically become squared under the sqrt (however Mathematicians may not like it stated that way).

 

[BigBeachBanana]

There are 3 problems

which one are you working with here?​

​

You do know that 3² = 9 and √9 = 3 ??

3 = √9 = √(3²) = 3

similarly

x = \(\displaystyle \sqrt{x^2} \)..........................correct expression would be →

|x| = \(\displaystyle \sqrt{x^2} \)

​

​

The OP says #21.

Hi everyone,

I'm finding calculus difficult...can someone please riddle me this? How does x^2 become x^2 in the denominator from the numerator and denominator being multiplied by 1/x?
I understand everything else, I think. Does x automatically become squared under the sqrt? That doesn't make sense.

Thank-you. Here's a screenshot. I'll talking about question #21.

View attachment 31800

I think you're asking about this specifically in the denominator:
[math]\sqrt{9x^2+x}\cdot \frac{1}{x}=\sqrt{9x^2+x}\cdot \sqrt{\left(\frac{1}{x}\right)^2}=\sqrt{9x^2+x}\cdot \sqrt{\frac{1}{x^2}}=\sqrt{(9x^2+x)\cdot\left(\frac{1}{x^2}\right)} =\sqrt{\frac{9x^2+x}{x^2}}[/math]The rest follows the solution.

 

[ScienceJen]

​

To answer your general question:

​

You do know that 3² = 9 and √9 = 3 ??

3 = √9 = √(3²) = 3

similarly

x = \(\displaystyle \sqrt{x^2} \)..........................correct expression would be →

|x| = \(\displaystyle \sqrt{x^2} \)

​

So to quote you - yes ... x (if it is a positive number) automatically become squared under the sqrt (however Mathematicians may not like it stated that way).

Hi there,

Yes, I know about sqrt 9 = 3.
I just recently learned that sqrt x²= absolute value x. .......................................................edited
Thank you.

The OP says #21.

I think you're asking about this specifically in the denominator:
[math]\sqrt{9x^2+x}\cdot \frac{1}{x}=\sqrt{9x^2+x}\cdot \sqrt{\left(\frac{1}{x}\right)^2}=\sqrt{9x^2+x}\cdot \sqrt{\frac{1}{x^2}}=\sqrt{(9x^2+x)\cdot\left(\frac{1}{x^2}\right)} =\sqrt{\frac{9x^2+x}{x^2}}[/math]The rest follows the solution.

Hi,
Yes, I am asking that specifically, thank you. Why are we not just dividing by the largest square root in the denominator for this exercise? Aka x^2.
Thanks again.

 

[Deleted member 4993]

that sqrt x = absolute value x.

I think you meant:

sqrt (x²) = absolute value x

 

[BigBeachBanana]

Hi there,

Yes, I know about sqrt 9 = 3.
I just recently learned that sqrt x = absolute value x.
Thank you.

Hi,
Yes, I am asking that specifically, thank you. Why are we not just dividing by the largest square root in the denominator for this exercise? Aka x^2.
Thanks again.

I don't understand your question. We are dividing by [imath]x^2[/imath].

 

[ScienceJen]

I think you meant:

sqrt (x²) = absolute value x

Yes, thank you

 

[ScienceJen]

Hi there,

Yes, I know about sqrt 9 = 3.
I just recently learned that sqrt x = absolute value x.
Thank you.

Hi,
Yes, I am asking that specifically, thank you. Why are we not just dividing by the largest square root in the denominator for this exercise? Aka x^2.
Thanks again.

Hi there,

Yes, I know about sqrt 9 = 3.
I just recently learned that sqrt x = absolute value x.
Thank you.

Hi,
Yes, I am asking that specifically, thank you. Why are we not just dividing by the largest square root in the denominator for this exercise? Aka x^2.
Thanks again.

Good morning,

I re-read the denominator solution again just now and it makes more sense. So, we divide the numerator and the denominator by 1/x in order to get rid of the x in the numerator. Then square x, so we can square root it to get it with the rest of the values under the square root? Don't we have to multiply the numerator and denominator by the same thing(s)? Just not sure how the rules work to go from 1/x to x^2.

Thank you for your help.

 

[ScienceJen]

I don't understand your question. We are dividing by [imath]x^2[/imath].

I guess I don't understand why we are multiplying by 1/x first.

 

[BigBeachBanana]

I guess I don't understand why we are multiplying by 1/x first.

If you were to do a direct substitution here, you'd get [imath]\frac{\infty}{\infty}[/imath], which is an indeterminate form.
A trick we deal with that is to divide the highest degree in both the numerator and the denominator, in this case, x.
That's what you're seeing here:

Note that [imath]\frac{1/x}{1/x}=1[/imath]. Essentially you're just multiplying by 1, which doesn't change the expression, however, it helps resolve the indeterminate issue. The rest is algebra, and you can see that when you do the direct substitution, you no longer have an indeterminate form, but a numerical answer.

 

[ScienceJen]

View attachment 31811
If you were to do a direct substitution here, you'd get [imath]\frac{\infty}{\infty}[/imath], which is an indeterminate form.
A trick we deal with that is to divide the highest degree in both the numerator and the denominator, in this case, x.
That's what you're seeing here:
View attachment 31810
Note that [imath]\frac{1/x}{1/x}=1[/imath]. Essentially you're just multiplying by 1, which doesn't change the expression, however, it helps resolve the indeterminate issue. The rest is algebra, and you can see that when you do the direct substitution, you no longer have an indeterminate form, but a numerical answer.
View attachment 31812

Great! Thank you so much!
I see now that sqrt of x^2 in the denominator is actually just x, essentially. Also, that 1/x is equivalent to dividing the numerator and denominator by x. I had trouble visualizing that.
I really appreciate your help.

 

[BigBeachBanana]

Great! Thank you so much!
I see now that sqrt of x^2 in the denominator is actually just x, essentially. Also, that 1/x is equivalent to dividing the numerator and denominator by x. I had trouble visualizing that.
I really appreciate your help.

I've shown you how x becomes x^2 in post #3.

 

[ScienceJen]

I've shown you how x becomes x^2 in post #3.

Yes, but I just "got it." I don't know if people who get math easily realize that for a lot of people, a concept needs to "click." At least for me. Then when I get it, I get it (usually).
Thank you!