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Let$ f(x) = \left\{ \begin{array}{ll} e^{-1/x^2} & \mbox{if} x \not= 0\\ 0 & \mbox{if} x = 0\\ \end{array} \right. $(a) Use the definition of derivative to compute $ f'(0) $.(b) Show that $ f $ has derivatives of all orders that are defined on $ \mathbb{R} $. [Hint: First show by induction that there is a polynomial $ p_n(x) $ and a nonnegative integer $ k_n $ such that $ f^{(n)}(x) = p_n(x)f(x)/x^{k_n} $ for $ x \not= 0 $.]

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(a) We show that $\lim _{x \rightarrow 0} \frac{f(x)}{x^{n}}=0$ for every integer $n \geq 0 .$ Let $y=\frac{1}{x^{2}} .$ Then\[\begin{array}{l}\lim _{x \rightarrow 0} \frac{f(x)}{x^{2 n}}=\lim _{x \rightarrow 0} \frac{e^{-1 / x^{2}}}{\left(x^{2}\right)^{n}}=\lim _{y \rightarrow \infty} \frac{y^{n}}{e^{y}} \stackrel{n}{=} \lim _{y \rightarrow \infty} \frac{n y^{n-1}}{e^{y}} \underline{H} \ldots=\underline{H} \lim _{y \rightarrow \infty} \frac{n !}{e^{y}}=0 \Rightarrow \\\lim _{x \rightarrow 0} \frac{f(x)}{x^{n}}=\lim _{x \rightarrow 0} x^{n} \frac{f(x)}{x^{2 n}}=\lim _{x \rightarrow 0} x^{n} \lim _{x \rightarrow 0} \frac{f(x)}{x^{2 n}}=0 . \text { Thus, } f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{f(x)}{x}=0\end{array}\](b) Using the Chain Rule and the Quotient Rule we see that $f^{(n)}(x)$ exists for $x \neq 0 .$ In fact, we prove by induction that foreach $n \geq 0,$ there is a polynomial $p_{n}$ and a non-negative integer $k_{n}$ with $f^{(n)}(x)=p_{n}(x) f(x) / x^{k_{n}}$ for $x \neq 0 .$ This istrue for $n=0 ;$ suppose it is true for the $n$ th derivative. Then $f^{\prime}(x)=f(x)\left(2 / x^{3}\right),$ so\[\begin{aligned}f^{(n+1)}(x) &=\left[x^{k_{n}}\left[p_{n}^{\prime}(x) f(x)+p_{n}(x) f^{\prime}(x)\right]-k_{n} x^{k_{n}-1} p_{n}(x) f(x)\right] x^{-2 k_{n}} \\&=\left[x^{k_{n}} p_{n}^{\prime}(x)+p_{n}(x)\left(2 / x^{3}\right)-k_{n} x^{k_{n}-1} p_{n}(x)\right] f(x) x^{-2 k_{n}} \\&=\left[x^{k_{n}+3} p_{n}^{\prime}(x)+2 p_{n}(x)-k_{n} x^{k_{n}+2} p_{n}(x)\right] f(x) x^{-\left(2 k_{n}+3\right)}\end{aligned}\] which has the desired form.Now we show by induction that $f^{(n)}(0)=0$ for all $n$. By part (a), $f^{\prime}(0)=0 .$ Suppose that $f^{(n)}(0)=0 .$ Then\[\begin{aligned}f^{(n+1)}(0) &=\lim _{x \rightarrow 0} \frac{f^{(n)}(x)-f^{(n)}(0)}{x-0}=\lim _{x \rightarrow 0} \frac{f^{(n)}(x)}{x}=\lim _{x \rightarrow 0} \frac{p_{n}(x) f(x) / x^{k_{n}}}{x}=\lim _{x \rightarrow 0} \frac{p_{n}(x) f(x)}{x^{k_{n}+1}} \\&=\lim _{x \rightarrow 0} p_{n}(x) \lim _{x \rightarrow 0} \frac{f(x)}{x^{k_{n}+1}}=p_{n}(0) \cdot 0=0\end{aligned}\]

15:46

Wen Zheng

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Derivatives

Differentiation

Volume

Harvey Mudd College

Baylor University

Idaho State University

Lectures

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for this problem, we want to use the definition of derivative. Um So substituting zero for X, we want to find f prime of zero. So based on this, we're eventually gonna have to use the hotel's rule. Um And that is because what we have is that f prime of zero equals the limit. As H approaches zero. F zero plus each -F of zero all over H. But since age is going to zero, what we see is we would get the indeterminant form. So we have to uh simplify this further and we'd eventually get it into the form H times. Uh so the limit is h approaches zero of H times. The limit as h approaches zero of 1 over E to the one over H squared. Since age is approaching zero, this would be eat to the infinity which is infinity. Yeah, But that makes this whole thing zero. And since H is approaching zero, this is also going to be zero. So the final answer will be zero as expected.

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