Question

find the limit please
\[\lim_{x \rightarrow infinity}\sqrt{x^2+3x+1}-x\]

Answer 1

you should put a 1 underneath that and rationalize the numerator by multiplying top and bottom by the conjugate of the top

Answer 2

\[\lim_{x \rightarrow \infty}\frac{\sqrt{x^2+3x+1}-x}{1} \cdot \frac{\sqrt{x^2+3x+1}+x}{\sqrt{x^2+3x+1}+x}\]

Answer 3

\[\lim_{x \rightarrow \infty}\frac{(x^2+3x+1)-x^2}{\sqrt{x^2+3x+1}+x}\]

Answer 4

\[\lim_{x \rightarrow \infty}\frac{3x+1}{\sqrt{x^2+3x+1}+x} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} \]

Answer 5

|x|=x if x>0 which it is since x gets positive large

Answer 6

\[\lim_{x \rightarrow \infty}\frac{3+\frac{1}{x}}{\sqrt{1+\frac{3}{x}+\frac{1}{x^2}}+\frac{x}{x}}\]

Answer 7

\[\frac{3+0}{\sqrt{1+0+0}+1}\]

Answer 8

thanks once again

Answer 9

lol np

Answer 10

this is the last one, can you help me with it please

Answer 11

Compute A= x^2+x/x+1 as x approaches +infinity and B= x^2+x/x^2+1 as x approaches -infinity

Answer 12

\[\lim_{x \rightarrow +\infty} (x^2+x) / (x+1) = +\infty\]\[\lim_{x \rightarrow -\infty}(x^2 +x)/(x^2+1) =1\]