Here's the question you clicked on:
kimmy0394
find limit as x approaches infinity sqrt. (x+3) / sqrt. (3x+1)
\[\lim_{x \rightarrow \infty}\frac{\sqrt{x+3}}{\sqrt{3x+1}}=\lim_{x \rightarrow \infty}\sqrt{\frac{x+3}{3x+1}}\]
now just look at the expression within the square root \[\lim_{x \rightarrow \infty}\frac{{x+3}}{{3x+1}}\] remember that it is all under a square root, so if the limit of that is negative, then there is no solution because you cant take the square root of a negative number
can you solve it now?
i thought it was 1/3 but it's not