Question

Find limit as x -> infinite, ((4 + (9x+1)^2)/(500x + 3x^2)

Answer 1

\[\lim_{x \rightarrow \infty} \frac{4+81x^2+18x+1}{500x+3x^2}\]
\[\lim_{x \rightarrow \infty} \frac{81x^2+18x+5}{3x^2+500x}=\frac{81}{3}=27\]

Answer 2

expand and simplify the numerator then

\[\lim_{x \rightarrow \infty}(81x^2 + 28x +5)/(500x + 3x^2)\]

then divide every term by x^2

will result in 81/3 as all other terms tend to zero as x approaches infinity

so the limit will be 27