Question

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

Short answer

The limit of the first function is 0, the limit of the second function is 1, and the limit of the third function is infinity.

Step-by-step solution

- Limit of the first function

To find \(\lim _{x \rightarrow \infty} \frac{x^{2}+2}{x^{3}-1}\), divide each term in the function by \(x^{3}\), the highest power of \(x\) in the denominator. The function simplifies to \(\lim _{x \rightarrow \infty} \frac{1/x+2/x^{3}}{1-1/x^{3}}\). Since as \(x\) approaches infinity, \(1/x\) and \(2/x^{3}\) both approach 0, the limit will be 0.

- Limit of the second function

Now, for \(\lim _{x \rightarrow \infty} \frac{x^{2}+2}{x^{2}-1}\), divide each term in the function by \(x^{2}\), the highest power of \(x\) in the denominator. The function simplifies to \(\lim _{x \rightarrow \infty} \frac{1+2/x^{2}}{1-1/x^{2}}\). Considering the limit as \(x\) approaches infinity, \(2/x^{2}\) and \(1/x^{2}\) both approach 0, thus the limit of this function is 1.

- Limit of the third function

Finally, for \(\lim _{x \rightarrow \infty} \frac{x^{2}+2}{x-1}\), since the power of \(x\) is higher in the numerator than in the denominator, the limit of the expression as \(x\) approaches infinity is infinity. Intuitively, as \(x\) grows larger, \(x^{2}\) increases much faster than \(x\), making the whole expression grow indefinitely.