Chapter 13.2, Problem 7E

To determine

To evaluate : The given limit.

Answer to Problem 7E

  $\underset{x\to -1}{\lim }\frac{x-2}{x^2+4x-3}=\frac{1}{2}$

Explanation of Solution

Given information:

  $\underset{x\to -1}{\lim }\frac{x-2}{x^2+4x-3}=?$

Formula used:

  $\underset{x\to a}{\lim }\left[f\left(x\right)+h\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }h\left(x\right)\text{ (Limit of a sum)}$

  $\underset{x\to a}{\lim }\left[f\left(x\right)-h\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)-\underset{x\to a}{\lim }h\left(x\right)\text{ (Limit of a Difference)}$

  $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)}\text{ if }\underset{x\to a}{\lim }g\left(x\right)\ne 0\text{ Limit of a Quotient}$

  $\underset{x\to a}{\lim }\left[cf\left(x\right)\right]=c\underset{x\to a}{\lim }f\left(x\right)\text{ Limit of a constant multiple}$

Some special limits

  $\underset{x\to a}{\lim }c=c$

  $\underset{x\to a}{\lim }x=a$

  $\underset{x\to a}{\lim }{x}^n=a^n\text{ Where }n\text{ is a positive integer}$

Calculation:

  $\begin{array}{c}\underset{x\to -1}{\lim }\frac{x-2}{x^2+4x-3}\\ =\frac{\underset{x\to -1}{\lim }\left(x-2\right)}{\underset{x\to -1}{\lim }\left(x^2+4x-3\right)}\text{ Limit of a Quotient}\\ =\frac{\underset{x\to -1}{\lim }x-\underset{x\to -1}{\lim }2}{\underset{x\to -1}{\lim }{x}^2+4\underset{x\to -1}{\lim }x-\underset{x\to -1}{\lim }3}\text{ Limit of a Sum, Difference and a constant multiple}\\ \text{=}\frac{\left(-1\right)-\left(2\right)}{{\left(-1\right)}^2+4\left(-1\right)-\left(3\right)}\text{ Special limits 1, 2 and 3}\\ \text{=}\frac{-3}{-6}\text{ Simplify}\\ =\frac{1}{2}\end{array}$