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no horizontal asymptote
a horizontal asymptote at y = 0
a horizontal asymptote at y = 1
\frac { x ^ { 3 } + 1 } { x ^ { 3 } + 2 } has

Explanation:

Step1: Find the limit as \(x\to\pm\infty\)

For a rational function \(\frac{f(x)}{g(x)}=\frac{x^{n}+...}{x^{m}+...}\), the horizontal - asymptote rule is: if \(n = m\), \(y=\frac{a_{n}}{b_{m}}\) (where \(a_{n}\) and \(b_{m}\) are the leading coefficients of \(f(x)\) and \(g(x)\)).
For the function \(y = \frac{x^{2}+1}{x^{2}+2}\), the degree of the numerator \(n = 2\) and the degree of the denominator \(m = 2\).
The leading coefficient of the numerator \(a_{n}=1\) and the leading coefficient of the denominator \(b_{m}=1\).

Step2: Calculate the limit

We use the formula \(\lim_{x\to\pm\infty}\frac{x^{2}+1}{x^{2}+2}=\lim_{x\to\pm\infty}\frac{x^{2}(1 + \frac{1}{x^{2}})}{x^{2}(1+\frac{2}{x^{2}})}\).
By the property \(\lim_{x\to\pm\infty}\frac{1}{x^{k}} = 0\) for \(k>0\), we have \(\lim_{x\to\pm\infty}\frac{1+\frac{1}{x^{2}}}{1 + \frac{2}{x^{2}}}\).
Substituting \(\lim_{x\to\pm\infty}\frac{1}{x^{2}}=0\) into the fraction, we get \(\frac{1 + 0}{1+0}=1\).

Answer:

A horizontal asymptote at \(y = 1\)