Let f be the function defined on R by f (0) = 0, f (x) = x2 sin 1/x for x≠0. The
value of D _ f (0), the lower left hand derivative of f at 0 is

To find the lower left-hand derivative of the function \( f \) at \( x = 0 \), we need to analyze the behavior of the function as \( x \) approaches 0 from the left. The function is defined as follows:

• \( f(0) = 0 \)
• \( f(x) = x^2 \sin(1/x) \) for \( x \neq 0 \)

The lower left-hand derivative, denoted as \( D_- f(0) \), is defined mathematically as:

\( D_- f(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} \)

Since \( f(0) = 0 \), we can simplify this to:

\( D_- f(0) = \lim_{h \to 0^-} \frac{f(h)}{h} \)

Now, substituting the expression for \( f(h) \) when \( h \neq 0 \), we have:

\( D_- f(0) = \lim_{h \to 0^-} \frac{h^2 \sin(1/h)}{h} \)

This simplifies to:

\( D_- f(0) = \lim_{h \to 0^-} h \sin(1/h) \)

Next, we need to evaluate the limit of \( h \sin(1/h) \) as \( h \) approaches 0 from the left. The sine function oscillates between -1 and 1, so we can bound \( \sin(1/h) \) as follows:

• \( -1 \leq \sin(1/h) \leq 1 \)

Multiplying through by \( h \) (which is negative as \( h \to 0^- \)), we get:

\( -h \leq h \sin(1/h) \leq h \)

As \( h \) approaches 0 from the left, both \( -h \) and \( h \) approach 0. By the Squeeze Theorem, we can conclude that:

\( \lim_{h \to 0^-} h \sin(1/h) = 0 \)

Thus, we find that:

\( D_- f(0) = 0 \)

In summary, the lower left-hand derivative of the function \( f \) at \( x = 0 \) is:

\( D_- f(0) = 0