Aufgabe 5 lim _(xarrow 4)(x^2-8x+16)/(2x-8)= f(x)=2x^3+sin(x)+(1)/(x) Longrightarrow f'(x)=(df)/(dx)(x)= int _(0)^1e^x+sqrt (x)dx= int (1)/(x^4)dx= __

1. $\lim _{x\rightarrow 4}\frac {x^{2}-8x+16}{2x-8}$<br /> = $\lim _{x\rightarrow 4}\frac {(x-4)^{2}}{2(x-4)}$<br /> = $\lim _{x\rightarrow 4}\frac {x-4}{2}$<br /> = $\frac {4-4}{2}$<br /> = 0<br /><br />2. $f(x)=2x^{3}+sin(x)+\frac {1}{x}$<br /> $\Longrightarrow f'(x)=\frac {df}{dx}(x)$<br /> = $6x^{2}+cos(x)-\frac {1}{x^{2}}$<br /><br />3. $\int _{0}^{1}e^{x}+\sqrt {x}dx$<br /> = $\int _{0}^{1}e^{x}dx+\int _{0}^{1}\sqrt {x}dx$<br /> = $[e^{x}]_{0}^{1}+\left[\frac {2}{3}x^{\frac {3}{2}}\right]_{0}^{1}$<br /> = $e-1+\frac {2}{3}$<br /> = $e-\frac {1}{3}$<br /><br />4. $\int \frac {1}{x^{4}}dx$<br /> = $\int x^{-4}dx$<br /> = $\frac {x^{-3}}{-3}+C$<br /> = $-\frac {1}{3x^{3}}+C$