Question

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

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

Short answer

Thus, the required answer is 8.

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,

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

Calculate the left hand limit of the given expression

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

\(\begin{array}{c}\mathop {\lim }\limits_{x \to - {4^ - }} \left( {\left| {x + 4} \right| - 2x} \right) &=& \mathop {\lim }\limits_{x \to - {4^ - }} \left( { - \left( {x + 4} \right) - 2x} \right)\\ &=& \mathop {\lim }\limits_{x \to - {4^ - }} \left( { - 3x - 4} \right)\\ &=& - 3\left( { - 4} \right) - 4\\ &=& 8\end{array}\)

Calculate the right hand limit of the given expression

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

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

As the left hand limit and right hand limit are equal, therefore the limits exist.

So, the value of the limit is 8.