Question

In Exercises \(11-14,\) find each limit, if possible. (a) \(\lim _{x \rightarrow \infty} \frac{5 x^{2}}{4 x^{2}+1}\) (b) \(\lim _{x \rightarrow \infty} \frac{5 x^{3 / 2}}{4 x^{3 / 2}+1}\) (c) \(\lim _{x \rightarrow \infty} \frac{5 x^{3 / 2}}{4 \sqrt{x}+1}\)

Short answer

(a) The limit as \(x\) approaches infinity is \(\frac{5}{4}\). (b) The limit as \(x\) approaches infinity is \(\frac{5}{4}\). (c) The limit as \(x\) approaches infinity is undefined.

Step-by-step solution

Find the limit of the first expression

Divide every term in the first expression \(\frac{5x^{2}}{4x^{2}+1}\) by \(x^{2}\):\n\(\frac{5x^{2}}{4x^{2}+1}\) becomes \(\frac{5}{4+\frac{1}{x^{2}}}\).\nAs \(x\) approaches infinity, \(\frac{1}{x^{2}}\) approaches 0 so the limit is \(\frac{5}{4}\)

Find the limit of the second expression

Divide every term in the second expression \(\frac{5x^{3/2}}{4x^{3/2}+1}\) by \(x^{3/2}\):\n\(\frac{5x^{3/2}}{4x^{3/2}+1}\) becomes \(\frac{5}{4+\frac{1}{x^{3/2}}}\).\nAs \(x\) approaches infinity, \(\frac{1}{x^{3/2}}\) approaches 0 so the limit is \(\frac{5}{4}\)

Find the limit of the third expression

Divide every term in the third expression \(\frac{5x^{3/2}}{4\sqrt{x}+1}\) by \(x^{3/2}\):\n\(\frac{5x^{3/2}}{4\sqrt{x}+1}\) becomes \(\frac{5}{\frac{4}{x^{1/2}}+\frac{1}{x^{3/2}}}\).\nAs \(x\) approaches infinity, both \(\frac{4}{x^{1/2}}\) and \(\frac{1}{x^{3/2}}\) approach 0 so the limit is \(5/0\), which is undefined.