Question

Evalauate the limit, if it exists.

\(\mathop {{\bf{lim}}}\limits_{x \to {\bf{6}}} \left( {{\bf{8}} - \frac{{\bf{1}}}{{\bf{2}}}x} \right)\)

Short answer

The value of the limit is 5.

Step-by-step solution

The limit laws

For the expression \(\mathop {\lim }\limits_{x \to 6} \left( {8 - \frac{1}{2}x} \right)\),Difference law is applicable.

According to the difference law, \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right)\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists.

Evaluate the limit

Solve the expression \(\mathop {\lim }\limits_{x \to 6} \left( {8 - \frac{1}{2}x} \right)\) as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 6} \left( {8 - \frac{1}{2}x} \right) &= \mathop {\lim }\limits_{x \to 6} 8 - \mathop {\lim }\limits_{x \to 6} \left( {\frac{x}{2}} \right)\\ &= 8 - \frac{1}{2}\left( 6 \right)\\ &= 8 - 3\\ &= 5\end{aligned}\)

So, the value of the limit is 5.