Question

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

Explanation:

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

Since \(- 1\leqslant\cos(\frac{1}{x})\leqslant1\) for \(x
eq0\).

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 \(\lim_{x
ightarrow0}\sqrt{x+\sqrt{x}}\).
Let \(t = \sqrt{x}\), then \(x=t^{2}\) and as \(x
ightarrow0\), \(t
ightarrow0\).
\(\lim_{x
ightarrow0}\sqrt{x+\sqrt{x}}=\lim_{t
ightarrow0}\sqrt{t^{2}+t}=\lim_{t
ightarrow0}\sqrt{t(t + 1)} = 0\).
Also, \(\lim_{x
ightarrow0}-\sqrt{x+\sqrt{x}}=0\).

Answer:

\(0\)