Question

1. $lim_{x

ightarrow+infty}\frac{1 + x^{2}}{1+e^{x}}$

Explanation:

Step1: Analyze the growth - rate of functions

As \(x\to+\infty\), the exponential function \(e^{x}\) grows much faster than the polynomial function \(x^{2}\).

Step2: Divide numerator and denominator by \(e^{x}\)

We have \(\lim_{x\to+\infty}\frac{1 + x^{2}}{1+e^{x}}=\lim_{x\to+\infty}\frac{\frac{1}{e^{x}}+\frac{x^{2}}{e^{x}}}{\frac{1}{e^{x}} + 1}\).

Step3: Use the limit rules

We know that \(\lim_{x\to+\infty}\frac{1}{e^{x}} = 0\) and \(\lim_{x\to+\infty}\frac{x^{n}}{e^{x}}=0\) for any positive integer \(n\). So \(\lim_{x\to+\infty}\frac{\frac{1}{e^{x}}+\frac{x^{2}}{e^{x}}}{\frac{1}{e^{x}} + 1}=\frac{0 + 0}{0 + 1}\).

Answer:

0