Question

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

Short answer

The solutions are: (a) 0, (b) -2/3, (c) undefined.

Step-by-step solution

Analyzing (a)

Let's analyze the function \(\lim _{x \rightarrow \infty} \frac{3-2 x}{3 x^{3}-1}\). If we plug \(x = \infty\) into the function, it will lead to an indeterminacy of the form 0/0. To resolve this indeterminacy, we will divide the numerator and the denominator by \(x^3\). So the function becomes \(\lim _{x \rightarrow \infty} \frac{3/x^{3}-2/x^{2}}{3 -1/x^{3}}\).

Solving (a)

Now take the limit as x approaches infinity. As \(x\) approaches infinity, \(3/x^{3}\), \(2/x^{2}\), and \(1/x^{3}\) each approach 0 because they are constants divided by a large number. Therefore, the function becomes \( \lim _{x \rightarrow \infty} \frac{-2/x^{2}}{3}\), which equals to 0.

Analyzing (b)

Now, let's analyze the function \(\lim _{x \rightarrow \infty} \frac{3-2 x}{3 x-1}\). If we plug \(x = \infty\) into the function, it will yet again lead to an indeterminacy of the form 0/0. To resolve this, we will divide the numerator and the denominator by \(x\). So the function becomes \(\lim _{x \rightarrow \infty} \frac{3/x -2}{3 -1/x}\).

Solving (b)

Now take the limit as x approaches infinity. As \(x\) approaches infinity, \(3/x\), and \(1/x\) each approach 0, hence, \( \lim _{x \rightarrow \infty} \frac{-2}{3}\), which equals to -2/3.

Analyzing (c)

Finally, let's analyze the function \(\lim _{x \rightarrow \infty} \frac{3-2 x^{2}}{3 x-1}\). If we plug \(x = \infty\) into the function, it results in an indeterminacy of the form 0/0. To resolve this, we divide the numerator and the denominator by \(x^2\). The function becomes \(\lim _{x \rightarrow \infty} \frac{3/x^{2} -2}{3/x -1/x^{2}}\).

Solving (c)

Now take the limit as x approaches infinity. As \(x\) approaches infinity, \(3/x^{2}\), \(3/x\), and \(1/x^{2}\) each approach 0, hence, \( \lim _{x \rightarrow \infty} \frac{-2}{0}\), which is undefined.