Exercise: $\newline$ $2022-2023$ $\newline$ $f(x)=\begin{cases} x-xe^{\frac{1}{x^{3}}}, & x > 0 \ \sin 2x, & x \leq 0 \end{cases}:$ $\newline$ a) Study continuity:

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Q. Exercise:

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2022-2023

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f(x)=\begin{cases} x-xe^{\frac{1}{x^{3}}}, & x > 0 \ \sin 2x, & x \leq 0 \end{cases}:

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a) Study continuity:

Check Continuity: To study the continuity of the piecewise function $f(x)$ , we need to check the continuity of each piece on its interval and then check the continuity at the point where the pieces meet, which is at $x = 0$ .
Function Analysis: First, let's consider the function $f(x) = x - xe^{\frac{1}{x^3}}$ for $x > 0$ . This function is a composition of continuous functions: the identity function $x$ , the exponential function $e^{\frac{1}{x^3}}$ , and the product of two functions. Since the composition of continuous functions is continuous, $f(x)$ is continuous for $x > 0$ .
Limit Calculation: Next, we consider the function $f(x) = \sin(2x)$ for $x \leq 0$ . The sine function is continuous everywhere, and multiplying its argument by $2$ does not affect its continuity. Therefore, $f(x)$ is continuous for $x \leq 0$ .
Evaluate Limit: Now, we need to check the continuity of $f(x)$ at $x = 0$ . For $f(x)$ to be continuous at $x = 0$ , the following must hold: $\newline$ $1$ . $f(0)$ is defined. $\newline$ $2$ . The limit of $f(x)$ as $x$ approaches $0$ from the left ( $\lim_{x\to0^-} f(x)$ ) equals $f(0)$ . $\newline$ $3$ . The limit of $f(x)$ as $x$ approaches $0$ from the right ( $x = 0$ $3$ ) equals $f(0)$ .
Apply L'Hôpital's Rule: First, we find $f(0)$ using the definition of $f(x)$ for $x \leq 0$ : $f(0) = \sin(2\cdot0) = \sin(0) = 0$ .So, $f(0)$ is defined and equals $0$ .
Transform Expression: Next, we find the limit of $f(x)$ as $x$ approaches $0$ from the left: $\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} \sin(2x) = \sin(0) = 0$ .
Final Continuity Check: Now, we find the limit of $f(x)$ as $x$ approaches $0$ from the right. This is more complicated because of the term $e^{1/x^3}$ . We need to evaluate: $\newline$ $\lim_{x\to 0^+} (x - xe^{1/x^3}).$ $\newline$ As $x$ approaches $0$ from the right, $x$ goes to $0$ and $e^{1/x^3}$ goes to infinity. However, the product of $x$ (which goes to $0$ ) and $e^{1/x^3}$ (which goes to infinity) is an indeterminate form $x$ $2$ . We need to apply L'Hôpital's rule or transform the expression to resolve this indeterminate form.
Final Continuity Check: Now, we find the limit of $f(x)$ as $x$ approaches $0$ from the right. This is more complicated because of the term $e^{1/x^3}$ . We need to evaluate: $\newline$ $\lim_{x\to0^+} (x - xe^{1/x^3}).$ $\newline$ As $x$ approaches $0$ from the right, $x$ goes to $0$ and $e^{1/x^3}$ goes to infinity. However, the product of $x$ (which goes to $0$ ) and $e^{1/x^3}$ (which goes to infinity) is an indeterminate form $x$ $2$ . We need to apply L'Hôpital's rule or transform the expression to resolve this indeterminate form.Let's transform the expression by multiplying and dividing by $x$ $3$ : $\newline$ $\lim_{x\to0^+} (x - xe^{1/x^3}) = \lim_{x\to0^+} (x - x^4 * (e^{1/x^3} / x^3)).$ $\newline$ Now, we can see that as $x$ approaches $0$ , $x$ $6$ approaches $0$ faster than $e^{1/x^3}$ grows, so the limit of $x$ $9$ as $x$ approaches $0$ is $0$ .
Final Continuity Check: Now, we find the limit of $f(x)$ as $x$ approaches $0$ from the right. This is more complicated because of the term $e^{(1/x^3)}$ . We need to evaluate: $\newline$ $\lim_{x\to0^+} (x - xe^{(1/x^3)}).$ $\newline$ As $x$ approaches $0$ from the right, $x$ goes to $0$ and $e^{(1/x^3)}$ goes to infinity. However, the product of $x$ (which goes to $0$ ) and $e^{(1/x^3)}$ (which goes to infinity) is an indeterminate form $x$ $2$ . We need to apply L'Hôpital's rule or transform the expression to resolve this indeterminate form.Let's transform the expression by multiplying and dividing by $x$ $3$ : $\newline$ $\lim_{x\to0^+} (x - xe^{(1/x^3)}) = \lim_{x\to0^+} (x - x^4 * (e^{(1/x^3)} / x^3)).$ $\newline$ Now, we can see that as $x$ approaches $0$ , $x$ $6$ approaches $0$ faster than $e^{(1/x^3)}$ grows, so the limit of $x$ $9$ as $x$ approaches $0$ is $0$ .Therefore, the limit of $f(x)$ as $x$ approaches $0$ from the right is: $\newline$ $\lim_{x\to0^+} (x - xe^{(1/x^3)}) = \lim_{x\to0^+} x - \lim_{x\to0^+} x^4 * (e^{(1/x^3)} / x^3) = 0 - 0 = 0.$
Final Continuity Check: Now, we find the limit of $f(x)$ as $x$ approaches $0$ from the right. This is more complicated because of the term $e^{(1/x^3)}$ . We need to evaluate: $\newline$ $\lim_{x\to0^+} (x - xe^{(1/x^3)})$ . $\newline$ As $x$ approaches $0$ from the right, $x$ goes to $0$ and $e^{(1/x^3)}$ goes to infinity. However, the product of $x$ (which goes to $0$ ) and $e^{(1/x^3)}$ (which goes to infinity) is an indeterminate form $x$ $2$ . We need to apply L'Hôpital's rule or transform the expression to resolve this indeterminate form.Let's transform the expression by multiplying and dividing by $x$ $3$ : $\newline$ $\lim_{x\to0^+} (x - xe^{(1/x^3)}) = \lim_{x\to0^+} (x - x^4 * (e^{(1/x^3)} / x^3))$ . $\newline$ Now, we can see that as $x$ approaches $0$ , $x$ $6$ approaches $0$ faster than $e^{(1/x^3)}$ grows, so the limit of $x$ $9$ as $x$ approaches $0$ is $0$ .Therefore, the limit of $f(x)$ as $x$ approaches $0$ from the right is: $\newline$ $\lim_{x\to0^+} (x - xe^{(1/x^3)}) = \lim_{x\to0^+} x - \lim_{x\to0^+} x^4 * (e^{(1/x^3)} / x^3) = 0 - 0 = 0$ .Since the limit from the left equals $0$ $6$ and the limit from the right equals $0$ $6$ , and $0$ $6$ is defined, $f(x)$ is continuous at $e^{(1/x^3)}$ $0$ .