- #1
pc2-brazil
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I study Calculus by myself, and I tried to solve the following question.
I got the answer, but is my solution consistent?
Thank you in advance.
1. The problem statement
Let f be a function such that |f(x)| ≤ x² for every x. Show that f is differentiable in 0 and that f'(0) = 0.
2. Relevant theorems
I used the Squeeze Theorem, as suggested by the book I'm using. This theorem states the following: "Suppose the functions f, g and h are defined in an open interval I containing a, except possibly at a, and that f(x) ≤ g(x) ≤ h(x) for every x in I such that x ≠ a. If the limits of both f(x) and h(x) as x approaches a exist and equal L, then the limit of g(x) as x approaches a also exists and equals L.".
If |f(x)| \leq x^2, then
-x^2 \leq f(x) \leq x^2 (1)
Derivative of f at 0:
f'(0)=\lim_{h\to 0}\frac{f(0+h) - f(0)}{h}
But, from (1), we see that f(0) = 0, so this derivative becomes:
f'(0)=\lim_{h\to 0}\frac{f(h)}{h} (2)
From (1), assuming h is approaching zero from positive values:
-h^2 \leq f(h) \leq h^2
-h \leq \frac{f(h)}{h} \leq h
From the Squeeze Theorem, we see that, since the limits of h and -h as h approaches 0 exist and equal 0, then
\lim_{h\to 0}\frac{f(h)}{h} = 0
Substituting this limit in (2), we see that f'(0)=0, as desired.
I got the answer, but is my solution consistent?
Thank you in advance.
1. The problem statement
Let f be a function such that |f(x)| ≤ x² for every x. Show that f is differentiable in 0 and that f'(0) = 0.
2. Relevant theorems
I used the Squeeze Theorem, as suggested by the book I'm using. This theorem states the following: "Suppose the functions f, g and h are defined in an open interval I containing a, except possibly at a, and that f(x) ≤ g(x) ≤ h(x) for every x in I such that x ≠ a. If the limits of both f(x) and h(x) as x approaches a exist and equal L, then the limit of g(x) as x approaches a also exists and equals L.".
The Attempt at a Solution
If |f(x)| \leq x^2, then
-x^2 \leq f(x) \leq x^2 (1)
Derivative of f at 0:
f'(0)=\lim_{h\to 0}\frac{f(0+h) - f(0)}{h}
But, from (1), we see that f(0) = 0, so this derivative becomes:
f'(0)=\lim_{h\to 0}\frac{f(h)}{h} (2)
From (1), assuming h is approaching zero from positive values:
-h^2 \leq f(h) \leq h^2
-h \leq \frac{f(h)}{h} \leq h
From the Squeeze Theorem, we see that, since the limits of h and -h as h approaches 0 exist and equal 0, then
\lim_{h\to 0}\frac{f(h)}{h} = 0
Substituting this limit in (2), we see that f'(0)=0, as desired.
