I copied a problem wrong, so i still need help!!!! LIMITS

[Professor Awesome]

i answered these problems but got them wrong and don't know why.

1) \(\displaystyle lim_x\rightarrow-\infty\) \(\displaystyle \frac {\sqrt{7+2x^4}} {3x+x^2}\)

i got \(\displaystyle +\infty\)

2) \(\displaystyle lim_x\rightarrow9^-\) \(\displaystyle \frac{9-x}{3-\sqrt{x}}\)

i got 6

3) \(\displaystyle lim_x\rightarrow-3^-\) \(\displaystyle \frac{-x}{x+3}\)

i have know idea how to do this one

 

[stapel]

Re: evaluating limits.

i answered these problems but got them wrong and don't know why.

Please show what you did and the answers you got. Thank you.

Eliz.

 

[Professor Awesome]

i did show the answers that i got
but for #1 I said it was infinity because the highest exponent is in the numerator, but i think it should be domething else because of that squareroot sign, but i don't know what it should be

for number two i did \(\displaystyle \frac{3+\sqrt{x}}{3+\sqrt{x}}\bullet\frac{9-x}{3-\sqrt{x}}\) I canceled and got \(\displaystyle 3+\sqrt{x}\) i plugged in 9 and got 6

i don't know how to do 3

 

[pka]

\(\displaystyle \frac{{\sqrt {7 + 2x^4 } }}{{3x + x^2 }} = \frac{{\left( {1/x^2 } \right)\sqrt {7 + 2x^4 } }}{{\left( {1/x^2 } \right)\left( {3x + x^2 } \right)}} = \frac{{\sqrt {\left( {7/x^4 } \right) + 2} }}{{\left( {3/x + 1} \right)}}\)

Now redo the limit in #1!

 

[Professor Awesome]

ok, what do I do with that?

 

[Professor Awesome]

i know that i have to learn this stuff on my own, but i got all of the other ones right except these 3 and pka I don't know what to do after that

 

[pka]

\(\displaystyle {{\rm{lim}}}\limits_{x \to \infty } \frac{{\sqrt {\left( {7/x^4 } \right) + 2} }}{{\left( {3/x + 1} \right)}} = \sqrt 2\)

 

[Professor Awesome]

thanks, but how did you get a 2 out of that mess?
and if you could just explain how to do the other two. you don't have to give an answer

 

[Professor Awesome]

oh shoot!!!!!!! the limit is approaching NEGATIVE INFINITY!!!!!!!!!!! SORRY

 

[Professor Awesome]

THESE ARE THE REAL QUESTIONS


1) \(\displaystyle lim_x\rightarrow-\infty\) \(\displaystyle \frac {\sqrt{7+2x^4}} {3x+x^2}\)

i got \(\displaystyle +\infty\)

2) \(\displaystyle lim_x\rightarrow9^-\) \(\displaystyle \frac{9-x}{3-\sqrt{x}}\)

i got 6

3) \(\displaystyle lim_x\rightarrow-3^-\) \(\displaystyle \frac{-x}{x+3}\)

i have know idea how to do this one

 

[stapel]

i did show the answers that i got

An hour after I asked you to post them, yes.

In the future, please post replies as replies, not as edits to previous posts. This better maintains the comprehensibility of the overall thread.

Thank you for your consideration.

Eliz.

 

[soroban]

Re: I copied a problem wrong, so i still need help!!!! LIMIT

Hello, Professor Awesome!

\(\displaystyle 1)\;\;\L\lim_{x\rightarrow-\infty}\frac {\sqrt{7\,+\,2x^4}} {3x\,+\,x^2}\)

Divide top and bottom by \(\displaystyle x^2\).


The numerator is: .\(\displaystyle \L\frac{\sqrt{7\,+\,2x^4}}{x^2}\:=\:\frac{\sqrt{7\,+\,2x^4}}{\sqrt{x^4}}\:=\:\sqrt{\frac{7\,+\,2x^4}{x^4}}\:=\:\sqrt{\frac{7}{x^4}\,+\,\frac{2x^4}{x^4}}\:=\:\sqrt{\frac{7}{x^4}\,+\,2}\)


The denominator is: .\(\displaystyle \L\frac{3x\,+\,x^2}{x^2}\:=\:\frac{3x}{x^2}\,+\,\frac{x^2}{x^2}\:=\:\frac{3}{x}\,+\,1\)


The limit is: .\(\displaystyle \L\lim_{x\to\infty}\left(\frac{\sqrt{\frac{7}{x^4}\,+\,2}}{\frac{3}{x}\,+\,1}\right)\;=\;\frac{\sqrt{0\,+\,2}}{0\,+\,1}\;=\;\sqrt{2}\)


[Edit: pka's too fast for me ... again!]