L’Hopital y formas indeterminadas

Ejercicio 54.1

Calcule cada uno de los límites siguientes e indique la forma indeterminada que se presenta.

1

\lim_{x \to 1} \frac{1 - x + \ln x}{x^{3} - 3 x + 2}

\frac{- 1}{6}

2

\lim_{x \to 0} \frac{2 sen (\frac{π}{4} - x) - \sqrt{2}}{x}

- \sqrt{2}

3

\lim_{v \to π ∕ 2} sen (\cos v) \sec v

1

4

\lim_{r \to 0^{+}} (e^{r} - 1) \ln r

0

5

\lim_{r \to π ∕ 2} r \tan r - \frac{π}{2} \sec r

- 1

6

\lim_{q \to 0^{+}} q^{(\frac{3}{4 + \ln q})}

e^{3}

7

\lim_{t \to {(π ∕ 2)}^{-}} {(\tan t)}^{2 t - π}

1

8

\lim_{𝜃 \to {(π ∕ 2)}^{-}} {(sen 𝜃)}^{\sec 𝜃}

1

9

\lim_{n \to \infty} {(1 + \frac{r}{n})}^{𝑛𝑡}

con

r

t

constantes,

t > 0

e^{𝑟𝑡}

10

\lim_{x \to 0} \frac{x e^{x} - x}{{sen}^{2} (2 x)}

\frac{1}{4}

11

\lim_{x \to 1} (\frac{1}{\ln x} - \frac{1}{x - 1})

\frac{1}{2}

12

\lim_{z \to + \infty} (z e^{1 ∕ z} - z)

1

13

\lim_{x \to {(π ∕ 2)}^{+}} {(x - π ∕ 2)}^{\cos x}

1

14

\lim_{x \to + \infty} {(\frac{x}{x + 1})}^{x}

e^{- 1}

15

\lim_{x \to 0} \frac{\ln (1 + x^{2})}{\cos (3 x) - e^{- x}}

0

16

\lim_{x \to 1} \frac{sen (x - 1)}{\ln (2 x - 1)}

\frac{1}{2}

17

\lim_{x \to 0} (arcsen x) (\csc x)

1

18

\lim_{x \to 0} (\frac{1}{x^{2}} - \frac{1}{x^{2} \sec x})

\frac{1}{2}

19

\lim_{x \to 0^{+}} x \ln (sen x)

0

20

\lim_{x \to 0} {(\cos (2 x))}^{1 ∕ x^{2}}

e^{- 2}

21

\lim_{x \to + \infty} {(1 + \frac{1}{2 x})}^{x^{2}}

+ \infty

22

\lim_{x \to 0} \frac{x - \tan x}{x - sen x}

- 2

23

\lim_{x \to 0} \frac{sen (𝑘𝑥) + \tan (𝑛𝑥)}{\arctan (𝑛𝑥)}, n \neq 0

\frac{k + n}{n}

24

\lim_{y \to 0} {(e^{2 y} - y)}^{y^{- 1}}

e

25

\lim_{x \to + \infty} \frac{- 1}{x^{2} {sen}^{2} (\frac{2}{x})}

\frac{- 1}{4}

26

\lim_{x \to + \infty} (x - 1) sen (\frac{𝜋𝑥}{x - 1})

- π

27

\lim_{x \to 0^{+}} \frac{e^{x} - (1 + x)}{x^{n}}, n = 1, 2, 3, . . .

a): Si n = 1 entonces $\lim_{x \to 0^{+}} \frac{e^{x} - (1 + x)}{x^{n}} = 0$ .
b): Si n = 2 entonces $\lim_{x \to 0^{+}} \frac{e^{x} - (1 + x)}{x^{n}} = \frac{1}{2}$ .
c): Si n $\geq 3$ entonces $\lim_{x \to 0^{+}} \frac{e^{x} - (1 + x)}{x^{n}} = + \infty$ .