# Let f(x) be a function satisfying |f(x)| ≤ x^2 for -1 ≤ x ≤ 1, how do you show that f is differentiable at x = 0 and find f’(0)?

The derivative exists and is zero.

Now

Thus

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To show that ( f(x) ) is differentiable at ( x = 0 ), we need to demonstrate that the limit ( \lim_{{h \to 0}} \frac{f(0 + h) - f(0)}{h} ) exists.

Given ( |f(x)| \leq x^2 ) for ( -1 \leq x \leq 1 ), we have ( -x^2 \leq f(x) \leq x^2 ). Since ( f(0) = 0 ), it follows that ( -x^2 \leq f(x) \leq x^2 ) for ( -1 \leq x \leq 1 ).

Now, considering ( \lim_{{h \to 0}} \frac{f(0 + h) - f(0)}{h} ), we have:

( \lim_{{h \to 0}} \frac{f(h)}{h} \leq \lim_{{h \to 0}} \frac{h^2}{h} = \lim_{{h \to 0}} h = 0 )

And

( \lim_{{h \to 0}} \frac{f(-h)}{h} \geq \lim_{{h \to 0}} \frac{-h^2}{h} = \lim_{{h \to 0}} -h = 0 )

Thus, both the left-hand and right-hand limits are equal to zero, which means ( f'(0) = 0 ).

Therefore, ( f(x) ) is differentiable at ( x = 0 ) and ( f'(0) = 0 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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