Question

find the limit of f(x) = \frac{5}{x}-2 as x approaches \infty and as x approaches -\infty.
lim f(x)
x→+∞
(type a simplified fraction.)

Explanation:

Step1: Recall limit rule for $\frac{1}{x}$

As $x
ightarrow\pm\infty$, $\lim_{x
ightarrow\pm\infty}\frac{1}{x}=0$.

Step2: Rewrite the function for limit calculation

We have $f(x)=\frac{5}{x}-2 = 5\times\frac{1}{x}-2$.

Step3: Calculate the limit as $x

ightarrow\infty$
Using the limit - sum rule $\lim_{x
ightarrow a}(f(x)+g(x))=\lim_{x
ightarrow a}f(x)+\lim_{x
ightarrow a}g(x)$ and the constant - multiple rule $\lim_{x
ightarrow a}(cf(x)) = c\lim_{x
ightarrow a}f(x)$, we get $\lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow\infty}(\frac{5}{x}-2)=\lim_{x
ightarrow\infty}\frac{5}{x}-\lim_{x
ightarrow\infty}2$. Since $\lim_{x
ightarrow\infty}\frac{5}{x}=5\lim_{x
ightarrow\infty}\frac{1}{x}=0$ and $\lim_{x
ightarrow\infty}2 = 2$, then $\lim_{x
ightarrow\infty}f(x)=0 - 2=-2$.

Step4: Calculate the limit as $x

ightarrow-\infty$
Similarly, $\lim_{x
ightarrow-\infty}f(x)=\lim_{x
ightarrow-\infty}(\frac{5}{x}-2)=\lim_{x
ightarrow-\infty}\frac{5}{x}-\lim_{x
ightarrow-\infty}2$. Since $\lim_{x
ightarrow-\infty}\frac{5}{x}=5\lim_{x
ightarrow-\infty}\frac{1}{x}=0$ and $\lim_{x
ightarrow-\infty}2 = 2$, then $\lim_{x
ightarrow-\infty}f(x)=0 - 2=-2$.

Answer:

-2