Question

lim_{x\to\infty}\frac{- 10x}{6 + 2x}=\square
lim_{x\to - \infty}\frac{5x - 3}{x^{3}+7x - 10}=\square
lim_{x\to\infty}\frac{x^{2}-13x - 14}{6 - 4x^{2}}=\square
lim_{x\to\infty}\frac{\sqrt{x^{2}+14x}}{3 - 13x}=\square
lim_{x\to - \infty}\frac{\sqrt{x^{2}+14x}}{3 - 13x}=\square

Explanation:

Step1: Divide by highest - power of x

For $\lim_{x
ightarrow\infty}\frac{- 10x}{6 + 2x}$, divide numerator and denominator by $x$: $\lim_{x
ightarrow\infty}\frac{-10}{\frac{6}{x}+2}$.
As $x
ightarrow\infty$, $\frac{6}{x}
ightarrow0$. So the limit is $\frac{-10}{0 + 2}=-5$.

Step2: Analyze degree of polynomials

For $\lim_{x
ightarrow-\infty}\frac{5x - 3}{x^{3}+7x - 10}$, the degree of the numerator is $n = 1$ and the degree of the denominator is $m=3$ with $m>n$. So $\lim_{x
ightarrow-\infty}\frac{5x - 3}{x^{3}+7x - 10}=0$.

Step3: Divide by highest - power of x for quadratic case

For $\lim_{x
ightarrow\infty}\frac{x^{2}-13x - 14}{6 - 4x^{2}}$, divide numerator and denominator by $x^{2}$: $\lim_{x
ightarrow\infty}\frac{1-\frac{13}{x}-\frac{14}{x^{2}}}{\frac{6}{x^{2}}-4}$.
As $x
ightarrow\infty$, $\frac{13}{x}
ightarrow0$, $\frac{14}{x^{2}}
ightarrow0$, $\frac{6}{x^{2}}
ightarrow0$. So the limit is $\frac{1-0 - 0}{0 - 4}=-\frac{1}{4}$.

Step4: Simplify square - root case for positive infinity

For $\lim_{x
ightarrow\infty}\frac{\sqrt{x^{2}+14x}}{3 - 13x}$, since $x>0$, $\sqrt{x^{2}+14x}=x\sqrt{1+\frac{14}{x}}$.
Divide numerator and denominator by $x$: $\lim_{x
ightarrow\infty}\frac{\sqrt{1+\frac{14}{x}}}{\frac{3}{x}-13}$.
As $x
ightarrow\infty$, $\frac{14}{x}
ightarrow0$, $\frac{3}{x}
ightarrow0$. So the limit is $\frac{\sqrt{1 + 0}}{0-13}=-\frac{1}{13}$.

Step5: Simplify square - root case for negative infinity

For $\lim_{x
ightarrow-\infty}\frac{\sqrt{x^{2}+14x}}{3 - 13x}$, since $x<0$, $\sqrt{x^{2}+14x}=-x\sqrt{1+\frac{14}{x}}$.
Divide numerator and denominator by $x$: $\lim_{x
ightarrow-\infty}\frac{-\sqrt{1+\frac{14}{x}}}{\frac{3}{x}-13}$.
As $x
ightarrow-\infty$, $\frac{14}{x}
ightarrow0$, $\frac{3}{x}
ightarrow0$. So the limit is $\frac{-\sqrt{1+0}}{0 - 13}=\frac{1}{13}$.

Answer:

-5
0
$-\frac{1}{4}$
$-\frac{1}{13}$
$\frac{1}{13}$