sesión para guardar los cambios en este archivo. Figura 2.10. Gráfica de la función \( y=\frac{x-9}{\sqrt{x}-3} \) Ejercicio 2 Encuentra los siguientes límites y da las gráficas de los ejercicios 5, 6 y 7 : 1. \( \lim _{x \rightarrow 3}\left(3 x^{2}-2 x-2\right) \) 6. \( \lim _{x \rightarrow 0} \frac{x^{7}-x^{5}}{x^{3}-x} \) 2. \( \lim _{x \rightarrow 0} \frac{x}{1+x^{2}} \) 7. \( \lim _{x \rightarrow 2} \frac{x^{3}-2 x-4}{x^{3}-8} \) 3. \( \lim _{x \rightarrow 2}(x-2)^{2} \) 8. \( \lim _{x \rightarrow 1} \frac{x^{3}-1}{x^{4}-1} \) 91 4. \( \lim _{x \rightarrow 3} \frac{\sqrt{x+6}-3}{x} \) 9. \( \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4} \) 5. \( \lim _{x \rightarrow 1} \frac{2 x^{3}+x^{2}-5 x+2}{x^{3}-1} \) 10. \( \lim _{x \rightarrow 4} \frac{x^{2}-16}{\sqrt{x}-2} \) Unidad 2 2.2.4. Límites infinitos En losejemplosanteriores setrabajó con indeterminaciones del tipo \( \frac{0}{0} \) quecomo sevio se resuel ven factorizando y cancelando el término lineal que las origina Ahora, ¿qué se hace si es sólo el denominador el que se anula en el valor en el que se quiere calcular el límite? Es decir, qué se hace en situaciones como la del siguiente ejemplo: Ejemplo 9 ¿Cuál es el \( \lim _{x \rightarrow 0} \frac{1}{x^{2}} \) ? Solución: si reemplazamos la x por el val or 0 obtenemos la expresión \( \frac{1}{0} \) que no es número real, yaque no existeningún número que multiplicado por 0 dé œomo resul tado el valor 1. Como estamos cal cul ando un límite, observemos lo que ocurre con los valores de la función cuando nos acercamos al valor 0 . En la tabla 2.2 aparecen al gunos val ores de \( x \) con sus correspondientes imágenes: Tabla 2.2. Valores de la función \( 1 / x^{2} \) en las cercanías de 0 .
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\( \lim _{x \rightarrow 3}\left(3 x^{2}-2 x-2\right) \) To find this limit, we can simply substitute the value 3 into the expression: \( \lim _{x \rightarrow 3}\left(3 x^{2}-2 x-2\right) = 3(3)^2 - 2(3) - 2 = 27 - 6 - 2 = 19 \) Show more…
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CHAPTER 3 Complex Differentiation 1. For each following function, determine the singular points at which the function is not analytic. Determine the derivative at the points where the function is analytic using Cauchy-Riemann Equations. (a) f(z) = 1 / (z + 1) (b) f(z) = z^2 + z + 1 (c) f(z) = |z|^2 2. Verify that the Cauchy-Riemann equations hold for the following functions and find their derivatives: (a) f(z) = z^3 (b) f(z) = e^z (c) f(z) = sin z 3. Suppose u(x, y) = x^2 - y^2. (a) Show that u is harmonic. (b) Find a harmonic conjugate function v(x, y). (c) Express f(z) in terms of z. 4. Show that f(z) = z* (the complex conjugate of z) is not differentiable anywhere. 5. Given f(z) = u + iv is analytic, show that both u and v satisfy Laplace's equation.
Manish J.
If you cannot make any headway with $\lim _{(x, y) \rightarrow(0,0)} f(x, y)$ in rectangular coordinates, try changing to polar coordinates. Substitute $x=r \cos \theta, y=r \sin \theta,$ and investigate the limit of the resulting expression as $r \rightarrow 0 .$ In other words, try to decide whether there exists a number $L$ satisfying the following criterion: Given $\varepsilon>0,$ there exists a $\delta>0$ such that for all $r$ and $\theta$ $$|r|<\delta \Rightarrow|f(r, \theta)-L|<\varepsilon $$If such an $L$ exists, then$$ \lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{r \rightarrow 0} f(r \cos \theta, r \sin \theta)=L$$ For instance,$$ \lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}}{x^{2}+y^{2}}=\lim _{r \rightarrow 0} \frac{r^{3} \cos ^{3} \theta}{r^{2}}=\lim _{r \rightarrow 0} r \cos ^{3} \theta=0$$ To verify the last of these equalities, we need to show that Equation (1) is satisfied with $f(r, \theta)=r \cos ^{3} \theta$ and $L=0 .$ That is, we need to show that given any $\varepsilon>0,$ there exists a $\delta>0$ such that for all $r$ and $\theta$ $$|r|<\delta \Rightarrow\left|r \cos ^{3} \theta-0\right|<\varepsilon$$ since $$ \left|r \cos ^{3} \theta\right|=|r|\left|\cos ^{3} \theta\right| \leq|r| \cdot 1=|r| $$the implication holds for all $r$ and $\theta$ if we take $\delta=\varepsilon$ In contrast, $$\frac{x^{2}}{x^{2}+y^{2}}=\frac{r^{2} \cos ^{2} \theta}{r^{2}}=\cos ^{2} \theta$$ takes on all values from 0 to 1 regardless of how small $|r|$ is, so that $\lim _{(x, y) \rightarrow(0.0)} x^{2} /\left(x^{2}+y^{2}\right)$ does not exist. In each of these instances, the existence or nonexistence of the limit as $r \rightarrow 0$ is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) $\theta=$ constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, $f(x, y)=\left(2 x^{2} y\right) /\left(x^{4}+y^{2}\right)$ becomes $$ f(r \cos \theta, r \sin \theta)=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\sin ^{2} \theta}$$ for $r \neq 0 .$ If we hold $\theta$ constant and let $r \rightarrow 0,$ the limit is $0 .$ On the path $y=x^{2},$ however, we have $r \sin \theta=r^{2} \cos ^{2} \theta$ and $$\begin{aligned} f(r \cos \theta, r \sin \theta) &=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\left(r \cos ^{2} \theta\right)^{2}} \\ &=\frac{2 r \cos ^{2} \theta \sin \theta}{2 r^{2} \cos ^{4} \theta}=\frac{r \sin \theta}{r^{2} \cos ^{2} \theta}=1 \end{aligned}$$Find the limit of $f$ as $(x, y) \rightarrow(0,0)$ or show that the limit does not exist.$$f(x, y)=\cos \left(\frac{x^{3}-y^{3}}{x^{2}+y^{2}}\right)$$
Partial Derivatives
Limits and Continuity in Higher Dimensions
If you cannot make any headway with $\lim _{(x, y) \rightarrow(0,0)} f(x, y)$ in rectangular coordinates, try changing to polar coordinates. Substitute $x=r \cos \theta, y=r \sin \theta,$ and investigate the limit of the resulting expression as $r \rightarrow 0 .$ In other words, try to decide whether there exists a number $L$ satisfying the following criterion: Given $\varepsilon>0,$ there exists a $\delta>0$ such that for all $r$ and $\theta$ $$|r|<\delta \Rightarrow|f(r, \theta)-L|<\varepsilon $$If such an $L$ exists, then$$ \lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{r \rightarrow 0} f(r \cos \theta, r \sin \theta)=L$$ For instance,$$ \lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}}{x^{2}+y^{2}}=\lim _{r \rightarrow 0} \frac{r^{3} \cos ^{3} \theta}{r^{2}}=\lim _{r \rightarrow 0} r \cos ^{3} \theta=0$$ To verify the last of these equalities, we need to show that Equation (1) is satisfied with $f(r, \theta)=r \cos ^{3} \theta$ and $L=0 .$ That is, we need to show that given any $\varepsilon>0,$ there exists a $\delta>0$ such that for all $r$ and $\theta$ $$|r|<\delta \Rightarrow\left|r \cos ^{3} \theta-0\right|<\varepsilon$$ since $$ \left|r \cos ^{3} \theta\right|=|r|\left|\cos ^{3} \theta\right| \leq|r| \cdot 1=|r| $$the implication holds for all $r$ and $\theta$ if we take $\delta=\varepsilon$ In contrast, $$\frac{x^{2}}{x^{2}+y^{2}}=\frac{r^{2} \cos ^{2} \theta}{r^{2}}=\cos ^{2} \theta$$ takes on all values from 0 to 1 regardless of how small $|r|$ is, so that $\lim _{(x, y) \rightarrow(0.0)} x^{2} /\left(x^{2}+y^{2}\right)$ does not exist. In each of these instances, the existence or nonexistence of the limit as $r \rightarrow 0$ is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) $\theta=$ constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, $f(x, y)=\left(2 x^{2} y\right) /\left(x^{4}+y^{2}\right)$ becomes $$ f(r \cos \theta, r \sin \theta)=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\sin ^{2} \theta}$$ for $r \neq 0 .$ If we hold $\theta$ constant and let $r \rightarrow 0,$ the limit is $0 .$ On the path $y=x^{2},$ however, we have $r \sin \theta=r^{2} \cos ^{2} \theta$ and $$\begin{aligned} f(r \cos \theta, r \sin \theta) &=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\left(r \cos ^{2} \theta\right)^{2}} \\ &=\frac{2 r \cos ^{2} \theta \sin \theta}{2 r^{2} \cos ^{4} \theta}=\frac{r \sin \theta}{r^{2} \cos ^{2} \theta}=1 \end{aligned}$$Find the limit of $f$ as $(x, y) \rightarrow(0,0)$ or show that the limit does not exist. $$f(x, y)=\frac{x^{3}-x y^{2}}{x^{2}+y^{2}}$$
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