(e) \\( lim _ { x \rightarrow 0 } sqrt { x + sqrt { x } } cos left( \frac { 1 } { x } \right) \\)

Explanation:

Step1: Analyze the range of (\cos(\frac{1}{x}))

Since (- 1\leqslant\cos(\frac{1}{x})\leqslant1) for (x\neq0).

Step2: Use the squeeze theorem

Multiply the inequality by (\sqrt{x+\sqrt{x}}) (note that (\sqrt{x+\sqrt{x}}\geqslant0) for (x\geqslant0)). We get (-\sqrt{x + \sqrt{x}}\leqslant\sqrt{x+\sqrt{x}}\cos(\frac{1}{x})\leqslant\sqrt{x+\sqrt{x}}).

Step3: Calculate the limits of the bounding functions

Calculate (\lim_{x\rightarrow0}\sqrt{x+\sqrt{x}}). Let (t = \sqrt{x}), then (x=t^{2}) and as (x\rightarrow0), (t\rightarrow0). (\lim_{x\rightarrow0}\sqrt{x+\sqrt{x}}=\lim_{t\rightarrow0}\sqrt{t^{2}+t}=\lim_{t\rightarrow0}\sqrt{t(t + 1)} = 0). Also, (\lim_{x\rightarrow0}-\sqrt{x+\sqrt{x}}=0).

Answer:

(0)