Answer
By Squeeze Theorem we obtain the limit:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,0} \right)} \left( {{x^2} - 16} \right)\cos \left( {\frac{1}{{{{\left( {x - 4} \right)}^2} + {y^2}}}} \right) = 0$
Work Step by Step
We have $ - 1 \le \cos \left( {\frac{1}{{{{\left( {x - 4} \right)}^2} + {y^2}}}} \right) \le 1$.
Suppose $x \ge 4$, so
$ - \left( {{x^2} - 16} \right) \le \left( {{x^2} - 16} \right)\cos \left( {\frac{1}{{{{\left( {x - 4} \right)}^2} + {y^2}}}} \right) \le {x^2} - 16$
Taking limits of the inequalities gives
$\mathop { - \lim }\limits_{\left( {x,y} \right) \to \left( {4,0} \right)} {x^2} - 16 \le \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,0} \right)} \left( {{x^2} - 16} \right)\cos \left( {\frac{1}{{{{\left( {x - 4} \right)}^2} + {y^2}}}} \right) \le \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,0} \right)} {x^2} - 16$
The two limits on both ends are equal to $0$.
Now, if $x < 4$, then
$\left( {{x^2} - 16} \right) \le \left( {{x^2} - 16} \right)\cos \left( {\frac{1}{{{{\left( {x - 4} \right)}^2} + {y^2}}}} \right) \le - \left( {{x^2} - 16} \right)$
Taking limits of the inequalities gives
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,0} \right)} {x^2} - 16 \le \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,0} \right)} \left( {{x^2} - 16} \right)\cos \left( {\frac{1}{{{{\left( {x - 4} \right)}^2} + {y^2}}}} \right) \le \mathop { - \lim }\limits_{\left( {x,y} \right) \to \left( {4,0} \right)} {x^2} - 16$
In this case too, the two limits on both ends are equal to $0$.
Therefore, by Squeeze Theorem we obtain the limit:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {4,0} \right)} \left( {{x^2} - 16} \right)\cos \left( {\frac{1}{{{{\left( {x - 4} \right)}^2} + {y^2}}}} \right) = 0$
