Question

find the limit.

$\underset{x\to -\infty }{\lim }\frac{3-x}{3+x}-2$

Short answer

The value of limit of the given function is $-3.$

Step-by-step solution

Step 1. Given information.

The limit of function is given here,

$\underset{x\to -\infty }{\lim }\frac{3-x}{3+x}-2$

Step 2. Formula used.

Here, in the given question we apply limit of a quotient and limit of sum, difference.

After it we apply the formula:

$\underset{x\to -\infty }{\lim }\frac{1}{x}=0$

$\underset{x\to -\infty }{\lim }\frac{1}{x^k}=0$

Step 3. Simplification.

Given,

$\begin{array}{l}\underset{x\to -\infty }{\lim }\frac{3-x}{3+x}-2\\ =\underset{x\to -\infty }{\lim }\frac{3-x-6-2x}{3+x}\end{array}$

Here, first we divide the numerator and denominator by the greatest power of role="math" localid="1648758985699" $x$given in the question then we apply limit of constant multiple.

So, divide the numerator and denominator by $x$.

$\begin{array}{c}\underset{x\to -\infty }{\lim }\frac{3-x-6-2x}{3+x}=\underset{x\to -\infty }{\lim }\left(\frac{\frac{3}{x}-1-\frac{6}{x}-2}{\frac{3}{x}+1}\right)\\ =\left(\frac{\underset{x\to -\infty }{\lim }\frac{3}{x}-\underset{x\to -\infty }{\lim }1-\underset{x\to -\infty }{\lim }\frac{6}{x}-\underset{x\to -\infty }{\lim }2}{\underset{x\to -\infty }{\lim }\frac{3}{x}+\underset{x\to -\infty }{\lim }1}\right)\end{array}$

Now, we apply the value of $x$which is equals to $-\infty .$

$\begin{array}{c}=\frac{0-1-0-2}{0+1}\\ =-3\end{array}$

Therefore, our final answer is $-3.$