Patel Fleming
01/28/2023 · Primary School
3. Utiliza tablas de valores para determinar el limite en cada caso. a. \( \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4} \) b. \( \lim _{x \rightarrow 2} \frac{x-2}{x^{2}+x-6} \) c. \( \lim _{x \rightarrow 1} \frac{x-1}{x^{3}-1} \) d. \( \lim _{x \rightarrow 0} \frac{e^{x}-1}{x} \) e. \( \lim _{x \rightarrow 0} \frac{\operatorname{sen}(x)}{x} \) f. \( \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x+4} \)
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Para determinar los límites utilizando tablas de valores, evaluaremos las funciones en puntos cercanos al valor al que se aproxima \( x \). A continuación, se presentan los cálculos para cada uno de los límites solicitados.
### a. \( \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4} \)
Evaluamos la función en valores cercanos a 4:
| \( x \) | \( \sqrt{x} - 2 \) | \( x - 4 \) | \( \frac{\sqrt{x}-2}{x-4} \) |
|----------|---------------------|--------------|-------------------------------|
| 3.9 | 0.974 | -0.1 | -9.74 |
| 3.99 | 0.995 | -0.01 | -99.5 |
| 4.01 | 1.005 | 0.01 | 100.5 |
| 4.1 | 1.05 | 0.1 | 10.5 |
A medida que \( x \) se aproxima a 4, el límite parece tender a \( \infty \) (positivo y negativo). Sin embargo, al aplicar la regla de L'Hôpital o simplificar, se puede encontrar que:
\[
\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} = \frac{1/2}{1} = \frac{1}{4}
\]
### b. \( \lim _{x \rightarrow 2} \frac{x-2}{x^{2}+x-6} \)
Evaluamos la función en valores cercanos a 2:
| \( x \) | \( x - 2 \) | \( x^2 + x - 6 \) | \( \frac{x-2}{x^2+x-6} \) |
|----------|--------------|---------------------|-----------------------------|
| 1.9 | -0.1 | -2.11 | 0.047 |
| 1.99 | -0.01 | -2.0101 | 0.0049 |
| 2.01 | 0.01 | -1.9899 | -0.005 |
| 2.1 | 0.1 | -1.1 | -0.0909 |
El límite se aproxima a 0. Por lo tanto:
\[
\lim_{x \to 2} \frac{x-2}{x^2+x-6} = 0
\]
### c. \( \lim _{x \rightarrow 1} \frac{x-1}{x^{3}-1} \)
Evaluamos la función en valores cercanos a 1:
| \( x \) | \( x - 1 \) | \( x^3 - 1 \) | \( \frac{x-1}{x^3-1} \) |
|----------|--------------|----------------|---------------------------|
| 0.9 | -0.1 | -0.271 | 0.369 |
| 0.99 | -0.01 | -0.0297 | 0.337 |
| 1.01 | 0.01 | 0.0297 | 0.337 |
| 1.1 | 0.1 | 0.271 | 0.369 |
El límite se aproxima a \( \frac{1}{3} \). Por lo tanto:
\[
\lim_{x \to 1} \frac{x-1}{x^3-1} = \frac{1}{3}
\]
### d. \( \lim _{x \rightarrow 0} \frac{e^{x}-1}{x} \)
Evaluamos la función en valores cercanos a 0:
| \( x \) | \( e^{x} - 1 \) | \( x \) | \( \frac{e^{x}-1}{x} \) |
|----------|------------------|---------|--------------------------|
| -0.1 | -0.0952 | -0.1 | 0.952 |
| -0.01 | -0.00995 | -0.01 | 0.995 |
| 0.01 | 0.00995 | 0.01 | 0.995 |
| 0.1 | 0.0952 | 0.1 | 0.952 |
El límite se aproxima a 1. Por lo tanto:
\[
\lim_{x \to 0} \frac{e^{x}-1}{x} = 1
\]
### e. \( \lim _{x \rightarrow 0} \frac{\operatorname{sen}(x)}{x} \)
Evaluamos la función en valores cercanos a 0:
| \( x \) | \( \sin(x) \) | \( x \) | \( \frac{\sin(x)}{x} \) |
|----------|----------------|---------|---------------------------|
| -0.1 | -0.09983 | -0.1 | 0.9983 |
| -0.01 | -0.00999983 | -0.01 | 0.999983 |
| 0.01 | 0.00999983 | 0.01 | 0.999983 |
| 0.1 | 0.09983 | 0.1 | 0.9983 |
El límite se aproxima a 1. Por lo tanto:
\[
\lim_{x \to 0} \frac{\sin(x)}{x} = 1
\]
### f. \( \lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x+4} \)
Evaluamos la función en valores cercanos a 4:
| \( x \) | \( \sqrt{x} - 2 \) | \( x + 4 \) | \( \frac{\sqrt{x}-2}{x+4} \) |
|----------|---------------------|--------------|-------------------------------|
| 3.9 | 0.974 | 7.9 | 0.123 |
| 3.99 | 0.995 | 7.99 | 0.124 |
| 4.01 | 1.005 | 8.01 | 0.125 |
| 4.1 | 1.05 | 8.1 | 0.129 |
El límite se aproxima a \( \frac{1}{8} \). Por lo tanto:
\[
\lim_{x \to 4} \frac{\sqrt{x}-2}{x+4} = \frac{1}{8}
\]
### Resumen de los límites:
a. \( \frac{1}{4} \)
b. \( 0 \)
c. \( \frac{1}{3} \)
d. \( 1 \)
e. \( 1 \)
f. \( \frac{1}{8} \)
Quick Answer
a. \( \frac{1}{4} \)
b. \( 0 \)
c. \( \frac{1}{3} \)
d. \( 1 \)
e. \( 1 \)
f. \( \frac{1}{8} \)
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