Question

evaluate the following limits. if needed, enter oo for ∞ and -oo for -∞. (a) \\(\lim_{x \to +\infty} (13x^2 + 10x^3) = \\) (b) \\(\lim_{x \to -\infty} (13x^2 + 10x^3) = \\) question help: video message instructor submit question jump to answer

Explanation:

Part (a)

Step 1: Analyze the leading term

For the polynomial \(13x^2 + 10x^3\), the leading term (the term with the highest power of \(x\)) is \(10x^3\) since the degree of \(x^3\) (which is 3) is higher than the degree of \(x^2\) (which is 2).

Step 2: Evaluate the limit as \(x \to \infty\)

As \(x\) approaches \(\infty\), the behavior of the polynomial is determined by its leading term. For the leading term \(10x^3\), when \(x\) becomes very large (approaches \(\infty\)), \(x^3\) also becomes very large (approaches \(\infty\)). Multiplying by 10 (a positive constant) still gives a value that approaches \(\infty\). So, \(\lim_{x \to \infty}(13x^2 + 10x^3)=\lim_{x \to \infty}10x^3=\infty\).

Part (b)

Step 1: Analyze the leading term

Again, the leading term is \(10x^3\).

Step 2: Evaluate the limit as \(x \to -\infty\)

When \(x\) approaches \(-\infty\), we consider the behavior of \(x^3\). For any real number \(x\), if \(x\) is negative, \(x^3\) is negative (because the cube of a negative number is negative: \((-a)^3=-a^3\) for \(a>0\)). So, when \(x\) is a large negative number (approaches \(-\infty\)), \(x^3\) is a large negative number (approaches \(-\infty\)). Multiplying by 10 (a positive constant) gives \(10x^3\) which approaches \(-\infty\). Therefore, \(\lim_{x \to -\infty}(13x^2 + 10x^3)=\lim_{x \to -\infty}10x^3 = -\infty\).

Answer:

s:
(a) \(\infty\)

(b) \(-\infty\)