Repeat the same argurnent to show that \[ \lim _{x \rightarrow \infty} \frac{10^{29} x}{x^{2}+4}=0 \text { and } \lim _{x \rightarrow \infty} \frac{x^{3}}{x^{2}+15}=\infty \]

To show that \(\lim _{x \rightarrow \infty} \frac{10^{29} x}{x^{2}+4}=0\), we can analyze the behavior of the function as \(x\) approaches infinity. As \(x\) approaches infinity, the term \(10^{29} x\) grows much faster than the term \(x^{2}+4\). This is because the exponent of \(x\) in the numerator is much larger than the exponent of \(x\) in the denominator. Therefore, the fraction \(\frac{10^{29} x}{x^{2}+4}\) approaches 0 as \(x\) approaches infinity. Similarly, to show that \(\lim _{x \rightarrow \infty} \frac{x^{3}}{x^{2}+15}=\infty\), we can observe that as \(x\) approaches infinity, the term \(x^{3}\) grows much faster than the term \(x^{2}+15\). This is because the exponent of \(x\) in the numerator is larger than the exponent of \(x\) in the denominator. Therefore, the fraction \(\frac{x^{3}}{x^{2}+15}\) approaches infinity as \(x\) approaches infinity. In summary, as \(x\) approaches infinity, the fraction \(\frac{10^{29} x}{x^{2}+4}\) approaches 0, and the fraction \(\frac{x^{3}}{x^{2}+15}\) approaches infinity.