Let IR denote the set of all real numbers. Define the function f: R→R by f(x)=\ [[2 - 2x ^ 2 - x ^ 2 * sin(1/x)], [2]] if z≠0, if x = 0 Find R.H.D and L.H.D

1. Function Definition

For $x \neq 0$:

For $x = 0$:

Notice that outside $x = 0$, the term $x^2$ multiplies everything, making the whole extra part very small near the origin.

2. Right‐Hand Derivative (RHD) Formula

Take $h \to 0^+$ and use

Similarly, for the left we use $h \to 0^-$. The key is to substitute $f(h)$, simplify, then squeeze the limit.

3. Squeezing Idea

Because $|\sin(1/h)| \le 1$ we have

Multiplying by $h$ makes

As $h \to 0$ (from either side) all three expressions go to $0$ by the Squeeze Theorem. Hence both RHD and LHD are $0$.

4. Why Both Sides Match

The only possible asymmetry would come from the sign of $h$ itself, yet the factor that could misbehave ($\sin(1/h)$) is still bounded. Multiplying a bounded number by tiny $h$ will always give a tiny number that tends to $0$, no matter the sign of $h$. Therefore, both directional limits coincide, and the full derivative at $0$ exists and is $0$.

What is the question?

We have a bumpy‐looking rule called a function. It tells us, “If you are at zero, I give you the number 2; if you are anywhere else, I give you a slightly smaller number that depends on $x$ in a wiggly way.”

We want to know how steep (the slope) the graph is exactly at $x = 0$, both from the right side and the left side.

How do we normally find a slope at a point?

Think of standing on a hill.
• Take a tiny step forward (right) and see how your height changes.
• Divide that tiny height change by the tiny step length.
• Shrink the step to almost zero.
That final number is the right‐hand derivative (RHD).
Do the same with a tiny backward step for the left‐hand derivative (LHD).

What happens in this problem?

When we plug in a super small number $h$ instead of $x$, the complicated wiggly part $h^2\sin\!\bigl(\tfrac{1}{h}\bigr)$ is still multiplied by $h^2$. Because $h^2$ is extra tiny, it kills the wiggles! The height change becomes so tiny that, after dividing by the tiny step, we still get something that goes to 0.

So from both sides, the slope flattens to 0.

Step–by–Step Solution

1. Write the difference quotient
   For $h \neq 0$,

   $$f(h) - f(0) \,=\, \Bigl[2 - 2h^2 - h^2\sin\!\bigl(\tfrac{1}{h}\bigr)\Bigr] - 2 \,=\, -h^2\bigl[2+\sin\!\bigl(\tfrac{1}{h}\bigr)\bigr].$$

2. Form RHD and LHD expressions

   $$\text{RHD} = \lim_{h \to 0^+} \frac{-h^2\bigl[2+\sin(1/h)\bigr]}{h} \,=\, \lim_{h \to 0^+} -h\bigl[2+\sin(1/h)\bigr].$$ $$\text{LHD} = \lim_{h \to 0^-} \frac{-h^2\bigl[2+\sin(1/h)\bigr]}{h} \,=\, \lim_{h \to 0^-} -h\bigl[2+\sin(1/h)\bigr].$$

3. Use the boundedness of $\sin$

   $$-1 \le \sin\!\left(\frac{1}{h}\right) \le 1 \;\;\Longrightarrow\;\; 1 \le 2+\sin\!\left(\frac{1}{h}\right) \le 3.$$

   Hence,

   $$|\, -h\bigl[2+\sin(1/h)\bigr] \,| \le 3|h|.$$

   Since $3|h| \to 0$ as $h \to 0$, by Squeeze Theorem both limits are $0$.

4. Conclusion

   $$\boxed{\text{RHD at }x=0 = 0} \quad \text{and}\quad \boxed{\text{LHD at }x=0 = 0}.$$

   Because RHD = LHD, the ordinary derivative at $x = 0$ also exists and equals $0$.