Question

Question 13IT, pre calc, I am at work so please answer so I can review it later tonight, thanks

Answer 1

EXPLANATION

Dividing the numerator and denominator by the highest denominator power (x^2):

$=\lim _{x\to\: -\infty\: }\mleft(\frac{\frac{1}{x}+\frac{1}{x^2}}{1-\frac{2}{x}}\mright)$

Applying the following property:

$\lim _{x\to a}\mleft[\frac{f\left(x\right)}{g\left(x\right)}\mright]=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)},\: \quad \lim _{x\to a}g\mleft(x\mright)\ne0$

$With\: the\: exception\: of\: indeterminate\: form$

$=\frac{\lim_{x\to\:-\infty\:}\left(\frac{1}{x}+\frac{1}{x^2}\right)}{\lim_{x\to\:-\infty\:}\left(1-\frac{2}{x}\right)}$

$=\frac{\lim_{x\to\: -\infty\: }(\frac{1}{x}+\frac{1}{x^2})}{\lim_{x\to\: -\infty\: }(1-\frac{2}{x})}=\frac{0}{1}=0$

Now, we need to apply the same steps to x-> ∞

$\mathrm{Apply\: the\: following\: algebraic\: property}\colon\quad a+b=a\mleft(1+\frac{b}{a}\mright)$

$\frac{x+1}{x^2-2x}=\frac{x\left(1+\frac{1}{x}\right)}{x^2\left(1-\frac{2}{x}\right)}$

$=\lim _{x\to\infty\: }\mleft(\frac{x\left(1+\frac{1}{x}\right)}{x^2\left(1-\frac{2}{x}\right)}\mright)$

Simplifying:

$=\lim _{x\to\infty\: }\mleft(\frac{1+\frac{1}{x}}{-2+x}\mright)$

$\lim _{x\to a}\mleft[\frac{f\left(x\right)}{g\left(x\right)}\mright]=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)},\: \quad \lim _{x\to a}g\mleft(x\mright)\ne0$

$\mathrm{With\: the\: exception\: of\: indeterminate\: form}$

$=\frac{\lim_{x\to\infty\:}\left(1+\frac{1}{x}\right)}{\lim_{x\to\infty\:}\left(-2+x\right)}$

$=\frac{1}{\infty\:}$

$\mathrm{Apply\: Infinity\: Property\colon}\: \frac{c}{\infty}=0$

$=0$

In conclusion, the appropiate end behavior is as follows:

$\lim _{x\to-\infty}f(x)=0;\text{ }lim_{x\to\infty}f(x)=0$