Question

Preciso da Resolucao da Conta (Limites)

Answer 1

5. $\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{f\left(-1+k \right )-f\left(-1 \right)}{k}$,  sendo $f\left(x \right )=11-3x^{2}$:

$\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{f\left(-1+k \right )-f\left(-1 \right)}{k}\\ \\ =\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{\left[\,11-3\cdot \left(1-k \right )^{2}\, \right ]-\left[\,11-3\cdot \left(-1 \right )^{2}\, \right ]}{k}\\ \\ =\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{\left[\,11-3\cdot \left(1-2k+k^{2} \right )\, \right ]-\left[\,11-3\cdot 1\, \right ]}{k}\\ \\ =\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{\left[\,11-3+6k-3k^{2}\, \right ]-\left[\,11-3\, \right ]}{k}\\ \\ =\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{\left[\,8+6k-3k^{2}\, \right ]-\left[\,8\, \right ]}{k}\\ \\ =\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{6k-3k^{2}}{k}\\ \\ =\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{k\cdot \left(6-3k \right )}{k}\\ \\ =\underset{k \to 0}{\mathrm{\ell im}}\;\left(6-3k \right )\\ \\ =6-3\cdot \left(0 \right )\\ \\ =6\\ \\ \\ \boxed{\underset{k \to 0}{\mathrm{\ell im}}\;\dfrac{f\left(-1+k \right )-f\left(-1 \right)}{k}=6}$


6. Se $f\left(x \right )=ax^{2}+bx+c$, então

$\underset{h \to 0}{\mathrm{\ell im}}\;\dfrac{f\left(x+h \right )-f\left(x \right )}{h}\\ \\ =\underset{h \to 0}{\mathrm{\ell im}}\;\dfrac{\left[\,a\cdot \left(x+h \right )^{2}+b\cdot \left(x+h \right )+c\, \right ]-\left[\,ax^{2}+bx+c\, \right ]}{h}\\ \\ =\underset{h \to 0}{\mathrm{\ell im}}\;\dfrac{\left[\,a\cdot \left(x^{2}+2xh+h^{2} \right )+b\cdot \left(x+h \right )+c\, \right ]-ax^{2}-bx-c}{h}\\ \\ =\underset{h \to 0}{\mathrm{\ell im}}\;\dfrac{\left[\,\diagup\!\!\!\!\!\! ax^{2}+2axh+ah^{2}+\diagup\!\!\!\!\!\! bx+bh+\diagup\!\!\!\! c\, \right ]-\diagup\!\!\!\!\!\! ax^{2}-\diagup\!\!\!\!\!\! bx-\diagup\!\!\!\! c}{h}\\ \\ =\underset{h \to 0}{\mathrm{\ell im}}\;\dfrac{2axh+ah^{2}+bh}{h}\\ \\ =\underset{h \to 0}{\mathrm{\ell im}}\;\dfrac{\diagup\!\!\!\! h\cdot \left(2ax+ah+b \right )}{\diagup\!\!\!\! h}\\ \\ =\underset{h \to 0}{\mathrm{\ell im}}\;\left(2ax+ah+b \right )\\ \\ =2ax+a\cdot \left(0 \right )+b\\ \\ =2ax+b$


$\boxed{\underset{h \to 0}{\mathrm{\ell im}}\;\dfrac{f\left(x+h \right )-f\left(x \right )}{h}=2ax+b}$