Question

1. what is the horizontal asymptote of the function $m(x)=\frac{x^{2}+ln(x)-cos(x)}{e^{2x}+x - 1}$?

Explanation:

Step1: Recall horizontal - asymptote rule

We find the limit as $x\to\pm\infty$. For large $x$, the dominant terms in the numerator and denominator are considered. As $x\to+\infty$, the dominant term in the numerator $y_1 = x^{2}+\ln(x)-\cos(x)$ is $x^{2}$ (since $\ln(x)$ grows slower than $x^{2}$ and $- 1\leqslant\cos(x)\leqslant1$ is bounded), and the dominant term in the denominator $y_2=e^{2x}+x - 1$ is $e^{2x}$.

Step2: Calculate the limit as $x\to+\infty$

We calculate $\lim_{x\to+\infty}\frac{x^{2}+\ln(x)-\cos(x)}{e^{2x}+x - 1}$. Using L'Hopital's rule multiple - times or knowing the growth rates of functions, we know that exponential functions grow faster than polynomial functions. The limit $\lim_{x\to+\infty}\frac{x^{2}}{e^{2x}}$. Let $u = 2x$, then $x=\frac{u}{2}$ and the limit becomes $\lim_{u\to+\infty}\frac{(\frac{u}{2})^{2}}{e^{u}}=\lim_{u\to+\infty}\frac{u^{2}}{4e^{u}}$.
Applying L'Hopital's rule once: $\lim_{u\to+\infty}\frac{2u}{4e^{u}}$, and applying it again: $\lim_{u\to+\infty}\frac{2}{4e^{u}} = 0$.
As $x\to-\infty$, $\ln(x)$ is not defined in the real - number system for $x\lt0$. So we only consider the limit as $x\to+\infty$.

Answer:

$y = 0$