- #1

TadeusPrastowo

- 21

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My https://www.amazon.com/dp/0073532320/?tag=pfamazon01-20 gives a rule of thumb to divide by the highest power in the

[tex]

\lim_{x\to\infty} \frac{4x^3+5}{-6x^2-7x} = \lim_{x\to\infty} \frac{4x+\frac{5}{x^2}}{-6-\frac{7}{x}}

[/tex]

But, the textbook gives

[tex]

\lim_{x\to\infty} \frac{4x^3+5}{-6x^2-7x} = \lim_{x\to\infty} \frac{x^3\left(4+\frac{5}{x^3}\right)}{-x^2\left(6+\frac{7}{x}\right)}

[/tex]

However, my alternative gives me ##\lim_{x\to\infty} -\frac{2}{3}x##. That is, a linear equation of ##-\frac{2}{3}x##. The textbook, however, resorts to performing a long division of ##4x^3+5## by ##-6x^2-7x## to obtain ##\lim_{x\to\infty} \left(-\frac{2}{3}x+\frac{7}{9}+\frac{5+\frac{49}{9}x}{-6x^2-7x}\right)##. That is, a linear equation of ##-\frac{2}{3}x+\frac{7}{9}##.

I have checked numerically that the slant asymptote is indeed the textbook's ##-\frac{2}{3}x+\frac{7}{9}## instead of my alternative's ##-\frac{2}{3}x##.

My question is then, why theoretically my alternative is wrong?

Thank you very much.

**denominator**for the following problem to demonstrate a slant (oblique) asymptote:[tex]

\lim_{x\to\infty} \frac{4x^3+5}{-6x^2-7x} = \lim_{x\to\infty} \frac{4x+\frac{5}{x^2}}{-6-\frac{7}{x}}

[/tex]

But, the textbook gives

**no explanation**whatsoever why it has to be the highest power in the denominator. As an alternative, I can actually do the following instead:[tex]

\lim_{x\to\infty} \frac{4x^3+5}{-6x^2-7x} = \lim_{x\to\infty} \frac{x^3\left(4+\frac{5}{x^3}\right)}{-x^2\left(6+\frac{7}{x}\right)}

[/tex]

However, my alternative gives me ##\lim_{x\to\infty} -\frac{2}{3}x##. That is, a linear equation of ##-\frac{2}{3}x##. The textbook, however, resorts to performing a long division of ##4x^3+5## by ##-6x^2-7x## to obtain ##\lim_{x\to\infty} \left(-\frac{2}{3}x+\frac{7}{9}+\frac{5+\frac{49}{9}x}{-6x^2-7x}\right)##. That is, a linear equation of ##-\frac{2}{3}x+\frac{7}{9}##.

I have checked numerically that the slant asymptote is indeed the textbook's ##-\frac{2}{3}x+\frac{7}{9}## instead of my alternative's ##-\frac{2}{3}x##.

My question is then, why theoretically my alternative is wrong?

Thank you very much.

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