Question

43-48 Find the limit, if it exists. If the limit does not exist, explain why.

44. \(\mathop {{\bf{lim}}}\limits_{x \to - {\bf{4}}} \frac{{\left| {x + {\bf{4}}} \right|}}{{{\bf{2}}x + {\bf{8}}}}\)

Short answer

Limit does not exist.

Step-by-step solution

Expand the expression given in limit using properties of modulus function

The function\(\left| {x + 4} \right|\) can be expanded as,

\(\begin{array}{c}\left| {x + 4} \right| &=& \left\{ {\begin{array}{*{20}{c}}{x + 4}&{{\rm{if}}\,\,x + 4 \ge 0}\\{ - \left( {x + 4} \right)}&{{\rm{if}}\,\,x + 4 < 0}\end{array}} \right.\\ &=& \left\{ {\begin{array}{*{20}{c}}{x + 4}&{{\rm{if}}\,\,x \ge - 4}\\{ - \left( {x + 4} \right)}&{{\rm{if}}\,\,x < - 4}\end{array}} \right.\end{array}\)

 Step 2: Calculate the left hand limit of the given expression

The left hand limit(\(x \to - {4^ - }\)) can be calculated as,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to - {4^ - }} \frac{{\left| {x + 4} \right|}}{{2x + 8}} &=& \mathop {\lim }\limits_{x \to - {4^ - }} \left( { - \frac{{\left( {x + 4} \right)}}{{2\left( {x + 4} \right)}}} \right)\\ &=& \mathop {\lim }\limits_{x \to - {4^ - }} \left( { - \frac{1}{2}} \right)\\ &=& - \frac{1}{2}\end{array}\)

Calculate the right hand limit of the given expression

The right hand limit (\(x \to - {4^ + }\)) can be calculated as,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to - {4^ + }} \left( {\frac{{\left| {x + 4} \right|}}{{2x + 8}}} \right) &=& \mathop {\lim }\limits_{x \to - {4^ + }} \left( {\frac{{x + 4}}{{2\left( {x + 4} \right)}}} \right)\\ &=& \mathop {\lim }\limits_{x \to - {4^ + }} \left( {\frac{1}{2}} \right)\\ &=& \frac{1}{2}\end{array}\)

As the left hand limit and right hand limit are not equal, therefore the limit does not exist.