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6. Let

g(x)=\sqrt{x} \sin (1 / x)

. a. Does

\lim _{x \rightarrow 0^{+}} g(x)

...

Oct 9, 2024

Solution by Steps

step 1

To find

\lim_{x \rightarrow 0^{+}} g(x)

, we use the Taylor series expansion:

g(0) + x g'(0) + \frac{1}{2} x^2 g''(0) + \frac{1}{6} g^{(3)}(0) x^3 + O(x^4)

. Since

g(0) = 0

, we need to evaluate the limit

step 2

The limit from the right is given by

\lim_{x \rightarrow 0^{+}} g(x) = \sqrt{x} \sin(1/x)

. As

x

approaches 0 from the right,

\sin(1/x)

oscillates between -1 and 1, thus

g(x)

oscillates between

-\sqrt{x}

and

\sqrt{x}

. Therefore, the limit does not exist

step 3

For

\lim_{x \rightarrow 0^{-}} g(x)

, we again consider the Taylor series. However,

g(x)

is not defined for x < 0 since

\sqrt{x}

is not real. Thus, the limit does not exist

step 4

Finally, for

\lim_{x \rightarrow 0} g(x)

, since the limits from the left and right do not exist, the overall limit does not exist

Answer

The limit

\lim_{x \rightarrow 0^{+}} g(x)

does not exist;

\lim_{x \rightarrow 0^{-}} g(x)

does not exist;

\lim_{x \rightarrow 0} g(x)

does not exist.

Key Concept

Limits and continuity of functions at points of oscillation

Explanation

The function

g(x) = \sqrt{x} \sin(1/x)

oscillates as

x

approaches 0, leading to non-existence of the limits from both sides.