Ejercicio 1
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Calcular, si existe, lim n → + ∞ ( 7 n 2 + 8 7 n 2 + 1 ) 6 n 2 + n \displaystyle \lim_{n \to +\infty} \left( \frac{7n^2 + 8}{7n^2 + 1} \right)^{6n^2 + n} n → + ∞ lim ( 7 n 2 + 1 7 n 2 + 8 ) 6 n 2 + n
Ver solución La base tiende a 1 1 1 + ∞ +\infty + ∞ 1 ∞ 1^\infty 1 ∞ 1 + ( algo que tiende a 0 ) 1 + (\text{algo que tiende a } 0) 1 + ( algo que tiende a 0 )
7 n 2 + 8 7 n 2 + 1 = 1 + 7 7 n 2 + 1 . \frac{7n^2 + 8}{7n^2 + 1} = 1 + \frac{7}{7n^2 + 1}. 7 n 2 + 1 7 n 2 + 8 = 1 + 7 n 2 + 1 7 . Entonces
( 1 + 7 7 n 2 + 1 ) 6 n 2 + n = [ ( 1 + 7 7 n 2 + 1 ) 7 n 2 + 1 7 ] 7 ( 6 n 2 + n ) 7 n 2 + 1 . \left( 1 + \frac{7}{7n^2 + 1} \right)^{6n^2 + n}
= \left[ \left( 1 + \frac{7}{7n^2 + 1} \right)^{\frac{7n^2 + 1}{7}} \right]^{\frac{7(6n^2 + n)}{7n^2 + 1}}. ( 1 + 7 n 2 + 1 7 ) 6 n 2 + n = ( 1 + 7 n 2 + 1 7 ) 7 7 n 2 + 1 7 n 2 + 1 7 ( 6 n 2 + n ) . Lo de adentro del corchete tiende a e e e n 2 n^2 n 2
lim n → + ∞ 42 n 2 + 7 n 7 n 2 + 1 = lim n → + ∞ 42 + 7 n 7 + 1 n 2 = 42 7 = 6. \lim_{n \to +\infty} \frac{42n^2 + 7n}{7n^2 + 1} = \lim_{n \to +\infty} \frac{42 + \frac{7}{n}}{7 + \frac{1}{n^2}} = \frac{42}{7} = 6. n → + ∞ lim 7 n 2 + 1 42 n 2 + 7 n = n → + ∞ lim 7 + n 2 1 42 + n 7 = 7 42 = 6. Por lo tanto el límite existe y vale
lim n → + ∞ ( 7 n 2 + 8 7 n 2 + 1 ) 6 n 2 + n = e 6 . \lim_{n \to +\infty} \left( \frac{7n^2 + 8}{7n^2 + 1} \right)^{6n^2 + n} = e^6. n → + ∞ lim ( 7 n 2 + 1 7 n 2 + 8 ) 6 n 2 + n = e 6 . Ejercicio 2
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Hallar a ∈ R a \in \mathbb{R} a ∈ R f : ( − 1 2 , 1 2 ) → R f : \left( -\frac{1}{2}, \frac{1}{2} \right) \to \mathbb{R} f : ( − 2 1 , 2 1 ) → R
f ( x ) = { ln ( x 2 + 1 ) 1 − cos ( x ) + 3 si x ≠ 0 , a si x = 0 f(x) = \begin{cases} \dfrac{\ln(x^2 + 1)}{1 - \cos(x)} + 3 & \text{si } x \neq 0, \\[2ex] a & \text{si } x = 0 \end{cases} f ( x ) = ⎩ ⎨ ⎧ 1 − cos ( x ) ln ( x 2 + 1 ) + 3 a si x = 0 , si x = 0 sea continua en x = 0 x = 0 x = 0
Ver solución Para que f f f 0 0 0 lim x → 0 f ( x ) = f ( 0 ) = a \displaystyle \lim_{x \to 0} f(x) = f(0) = a x → 0 lim f ( x ) = f ( 0 ) = a
El cociente ln ( x 2 + 1 ) 1 − cos ( x ) \dfrac{\ln(x^2 + 1)}{1 - \cos(x)} 1 − cos ( x ) ln ( x 2 + 1 ) 0 0 \frac{0}{0} 0 0
lim u → 0 ln ( 1 + u ) u = 1 , lim x → 0 1 − cos ( x ) x 2 = 1 2 . \lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1, \qquad \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2}. u → 0 lim u ln ( 1 + u ) = 1 , x → 0 lim x 2 1 − cos ( x ) = 2 1 . Multiplicamos y dividimos por x 2 x^2 x 2
ln ( x 2 + 1 ) 1 − cos ( x ) = ln ( 1 + x 2 ) x 2 ⋅ x 2 1 − cos ( x ) → x → 0 1 ⋅ 2 = 2. \frac{\ln(x^2 + 1)}{1 - \cos(x)} = \frac{\ln(1 + x^2)}{x^2} \cdot \frac{x^2}{1 - \cos(x)} \xrightarrow[x \to 0]{} 1 \cdot 2 = 2. 1 − cos ( x ) ln ( x 2 + 1 ) = x 2 ln ( 1 + x 2 ) ⋅ 1 − cos ( x ) x 2 x → 0 1 ⋅ 2 = 2. (En el primer factor usamos u = x 2 → 0 u = x^2 \to 0 u = x 2 → 0
Entonces
lim x → 0 f ( x ) = 2 + 3 = 5 , \lim_{x \to 0} f(x) = 2 + 3 = 5, x → 0 lim f ( x ) = 2 + 3 = 5 , y f f f x = 0 x = 0 x = 0 a = 5 a = 5 a = 5
Ejercicio 3
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Encontrar el dominio y la imagen de f ( x ) = 4 x e 5 − x 2 f(x) = 4\sqrt{x}\, e^{5 - x^2} f ( x ) = 4 x e 5 − x 2
Ver solución Dominio. La exponencial está definida para todo x x x x ≥ 0 x \geq 0 x ≥ 0 Dom ( f ) = [ 0 , + ∞ ) \operatorname{Dom}(f) = [0, +\infty) Dom ( f ) = [ 0 , + ∞ )
Imagen. Como x ≥ 0 \sqrt{x} \geq 0 x ≥ 0 e 5 − x 2 > 0 e^{5 - x^2} > 0 e 5 − x 2 > 0 f ( x ) ≥ 0 f(x) \geq 0 f ( x ) ≥ 0 f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0
Derivamos (para x > 0 x > 0 x > 0
f ′ ( x ) = 4 e 5 − x 2 ( 1 2 x + x ⋅ ( − 2 x ) ) = 4 e 5 − x 2 ⋅ 1 − 4 x 2 2 x . f'(x) = 4 e^{5 - x^2} \left( \frac{1}{2\sqrt{x}} + \sqrt{x} \cdot (-2x) \right)
= 4 e^{5 - x^2} \cdot \frac{1 - 4x^2}{2\sqrt{x}}. f ′ ( x ) = 4 e 5 − x 2 ( 2 x 1 + x ⋅ ( − 2 x ) ) = 4 e 5 − x 2 ⋅ 2 x 1 − 4 x 2 . El signo de f ′ f' f ′ 1 − 4 x 2 1 - 4x^2 1 − 4 x 2 ( 0 , 1 2 ) (0, \frac{1}{2}) ( 0 , 2 1 ) f f f ( 1 2 , + ∞ ) (\frac{1}{2}, +\infty) ( 2 1 , + ∞ ) f f f x = 1 2 x = \frac{1}{2} x = 2 1
f ( 1 2 ) = 4 ⋅ 1 2 ⋅ e 5 − 1 4 = 2 2 e 19 / 4 . f\left( \tfrac{1}{2} \right) = 4 \cdot \frac{1}{\sqrt{2}} \cdot e^{5 - \frac{1}{4}} = 2\sqrt{2}\, e^{19/4}. f ( 2 1 ) = 4 ⋅ 2 1 ⋅ e 5 − 4 1 = 2 2 e 19/4 . Además, lim x → + ∞ 4 x e 5 − x 2 = lim x → + ∞ 4 e 5 x e x 2 = 0 \displaystyle \lim_{x \to +\infty} 4\sqrt{x}\, e^{5 - x^2} = \lim_{x \to +\infty} \frac{4 e^5 \sqrt{x}}{e^{x^2}} = 0 x → + ∞ lim 4 x e 5 − x 2 = x → + ∞ lim e x 2 4 e 5 x = 0
Como f f f [ 0 , 1 2 ] [0, \frac{1}{2}] [ 0 , 2 1 ] 0 0 0 2 2 e 19 / 4 2\sqrt{2}\, e^{19/4} 2 2 e 19/4 [ 1 2 , + ∞ ) [\frac{1}{2}, +\infty) [ 2 1 , + ∞ ) 0 0 0
Im ( f ) = [ 0 , 2 2 e 19 / 4 ] . \operatorname{Im}(f) = \left[ 0,\ 2\sqrt{2}\, e^{19/4} \right]. Im ( f ) = [ 0 , 2 2 e 19/4 ] . Ejercicio 4
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Sea f : [ 1 , 9 ] → R f : [1, 9] \to \mathbb{R} f : [ 1 , 9 ] → R f ( x ) = 4 x 4 − 24 x 6 f(x) = \dfrac{4}{x^4} - \dfrac{24}{x^6} f ( x ) = x 4 4 − x 6 24 f f f [ 1 , 9 ] [1, 9] [ 1 , 9 ]
Ver solución f f f [ 1 , 9 ] [1, 9] [ 1 , 9 ]
Puntos críticos. Escribimos f ( x ) = 4 x − 4 − 24 x − 6 f(x) = 4x^{-4} - 24x^{-6} f ( x ) = 4 x − 4 − 24 x − 6
f ′ ( x ) = − 16 x − 5 + 144 x − 7 = 144 − 16 x 2 x 7 = 16 ( 9 − x 2 ) x 7 . f'(x) = -16x^{-5} + 144x^{-7} = \frac{144 - 16x^2}{x^7} = \frac{16(9 - x^2)}{x^7}. f ′ ( x ) = − 16 x − 5 + 144 x − 7 = x 7 144 − 16 x 2 = x 7 16 ( 9 − x 2 ) . En [ 1 , 9 ] [1, 9] [ 1 , 9 ] f ′ ( x ) = 0 ⟺ x 2 = 9 ⟺ x = 3 f'(x) = 0 \iff x^2 = 9 \iff x = 3 f ′ ( x ) = 0 ⟺ x 2 = 9 ⟺ x = 3
Evaluamos:
f ( 1 ) = 4 − 24 = − 20 , f ( 3 ) = 4 81 − 24 729 = 36 − 24 729 = 12 729 = 4 243 , f ( 9 ) = 4 6561 − 24 531441 = 324 − 24 531441 = 300 531441 = 100 177147 . \begin{aligned}
f(1) &= 4 - 24 = -20, \\
f(3) &= \frac{4}{81} - \frac{24}{729} = \frac{36 - 24}{729} = \frac{12}{729} = \frac{4}{243}, \\
f(9) &= \frac{4}{6561} - \frac{24}{531441} = \frac{324 - 24}{531441} = \frac{300}{531441} = \frac{100}{177147}.
\end{aligned} f ( 1 ) f ( 3 ) f ( 9 ) = 4 − 24 = − 20 , = 81 4 − 729 24 = 729 36 − 24 = 729 12 = 243 4 , = 6561 4 − 531441 24 = 531441 324 − 24 = 531441 300 = 177147 100 . Como 4 243 ≈ 0,0165 \frac{4}{243} \approx 0{,}0165 243 4 ≈ 0 , 0165 100 177147 ≈ 0,00056 \frac{100}{177147} \approx 0{,}00056 177147 100 ≈ 0 , 00056
El máximo absoluto es 4 243 \frac{4}{243} 243 4 x = 3 x = 3 x = 3
El mínimo absoluto es − 20 -20 − 20 x = 1 x = 1 x = 1