anonymous
  • anonymous
find the limit please \[\lim_{x \rightarrow infinity}\sqrt{x^2+3x+1}-x\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
you should put a 1 underneath that and rationalize the numerator by multiplying top and bottom by the conjugate of the top
myininaya
  • myininaya
\[\lim_{x \rightarrow \infty}\frac{\sqrt{x^2+3x+1}-x}{1} \cdot \frac{\sqrt{x^2+3x+1}+x}{\sqrt{x^2+3x+1}+x}\]
myininaya
  • myininaya
\[\lim_{x \rightarrow \infty}\frac{(x^2+3x+1)-x^2}{\sqrt{x^2+3x+1}+x}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

myininaya
  • myininaya
\[\lim_{x \rightarrow \infty}\frac{3x+1}{\sqrt{x^2+3x+1}+x} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} \]
myininaya
  • myininaya
|x|=x if x>0 which it is since x gets positive large
myininaya
  • myininaya
\[\lim_{x \rightarrow \infty}\frac{3+\frac{1}{x}}{\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+\frac{x}{x}}\]
myininaya
  • myininaya
\[\frac{3+0}{\sqrt{1+0+0}+1}\]
anonymous
  • anonymous
thanks once again
myininaya
  • myininaya
lol np
anonymous
  • anonymous
this is the last one, can you help me with it please
anonymous
  • anonymous
Compute A= x^2+x/x+1 as x approaches +infinity and B= x^2+x/x^2+1 as x approaches -infinity
anonymous
  • anonymous
\[\lim_{x \rightarrow +\infty} (x^2+x) / (x+1) = +\infty\]\[\lim_{x \rightarrow -\infty}(x^2 +x)/(x^2+1) =1\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.