MAT01353 - CÁLCULO E GEOMETRIA ANALÍTICA I-A - UFRGS
EXERCÍCIOS DE PRÉ-CÁLCULO
1. As expressões abaixo foram retiradas de problemas de limite (do tipo $\displaystyle 0/0$), da disciplina de Cálculo I. Classifique o desenvolvimento apresentado em cada alternativa como falso (F) ou verdadeiro (V), conforme possuir erros matemáticos ou não.
(a) (F)(V)$\displaystyle \frac{x^2+3x}{x^2-9} = \frac{x \cancel{(x+3)}}{\cancel{(x+3)}(x-3)} = \frac{x}{x-3}$
(b) (F)(V) $\displaystyle \frac{x^2 + x}{x^2 - 1} = \frac{x\cancel{(x+1)}}{(x-1)\cancel{(x+1)}} = \frac{x}{x-1} = 1 - x$
(c) (F)(V)$\displaystyle \frac{x^3-2x^2}{x^2-4} = x + \frac{x^2}{2} = x \left( 1 + \frac{x}{2} \right) $
(d) (F)(V)$\displaystyle \frac{x^3-3x-2}{x^2-4} = \fra{(x^2+2x+1)\cancel{(x-2)}}{(x+2)\cancel{(x-2)}} = \frac{x^2 + 2x + 1}{x+2}$
(e) (F)(V)$\displaystyle \frac{\sqrt{4+h}-2}{h} = \frac{\bcancel{2}+\sqrt{h}-\bcancel{2}}{h} = \frac{1}{\sqrt{h}}$
(f) (F)(V)$\displaystyle \frac{\sqrt[3]{1+h}-1}{h} = \frac{\sqrt{(1+h)^3}-1}{h} = \frac{\bcancel{1}+h^{3/2}-\bcancel{1}}{h} = \sqrt{h}$
(g) (F)(V)$\displaystyle \frac{\sqrt{1+h^3}-1}{h} = \frac{\sqrt{1+h^3}-1}{h} \cdot \frac{... ...ancel{1}+h^3 - \bcancel{1}}{h (\sqrt{1+h^3}+1) } = \frac{h^2}{1 +\sqrt{1+h^3}} $
(h) (F)(V)$\displaystyle \frac{(x+h)^3-x^3}{h} = \frac{y^3-x^3}{y-x} = y^2 - x^2$, onde $y = x+h$
(i) (F)(V)
$\displaystyle \frac{(\pi+h)\cos(\pi+h)+\pi}{h} = \frac{(\pi+h)(-1 + \cos(h)) + \pi}{h} $
$\displaystyle = \frac{\bcancel{-\pi} - h + (\pi+h) \cos(h) + \bcancel{\pi}}{h} = (1 + \pi/h) \cos(h) -1 \ , \ h \neq 0 $
(j) (F)(V) $\displaystyle \frac{1}{1+h^2}-1= \frac{\cancel{1} - (\cancel{1}+h^2)}{1+h^2} = \frac{-h^2}{1+h^2}$
(k) (F)(V) $\displaystyle \frac{e^{x^2}-1}{x} = \frac{(e^x-1)(e^x+1)}{x} = \frac{e^x-1}{x} \cdot (e^x + 1)$
(ℓ) (F)(V) $\displaystyle \frac{e^{3x}-1}{x}= \frac{(u-1)(u^2 + u + 1)}{x}$, onde $u = e^x$
(m) (F)(V) $\displaystyle \frac{\ln(1+h^3)}{h} = \ln\left( 1/h + h^2 \right) = \ln(-h + h^2) = \ln( h(h-1) ) = \ln(h)-\ln(h-1)$
(n) (F)(V) $\displaystyle \frac{\ln(1+e^h}{h^2} = \frac{\ln(1+e^h)}{\ln(e^{h^2}} = \ln(1+e^h) - \ln\left(e^{h^2}\right)$
(o) (F)(V) $\displaystyle \frac{\cos(\pi+h)-\cos(\pi)}{h} = \frac{\cos(\pi)\cos(h)+1}{h} = \frac{1-\cos(h)}{h}$
(p) (F)(V)$\displaystyle \frac{x}{x+3} = \frac{x+3-3}{x+3} = 1 - \frac{3}{x+3} \ , \ x \neq -3 $
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JBC, 15/04/13,19/5/13