Question

find the limit.

$\underset{x\to -\infty }{\lim }\left(\frac{x-1}{x+1}+6\right)$

Short answer

The value of limit of the given function is $7.$

Step-by-step solution

Step 1. Given information.

The limit of function is given here,

$\underset{x\to -\infty }{\lim }\left(\frac{x-1}{x+1}+6\right)$

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,

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

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

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

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

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

$\begin{array}{c}=\frac{1-0+6+0}{1+0}\\ =1+6\\ =7\end{array}$

Therefore, our final answer is $7.$