Comprehensive Exercises on Important Limits

Approach: Use the generalized form of the first important limit.

Steps:

1. $\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin 5x}{5x}$

2. $= 5 \cdot \lim_{x \to 0} \frac{\sin 5x}{5x} = 5 \cdot 1 = 5$

Answer: The limit is 5.

Approach: Use the generalized form of the second important limit.

Steps:

1. $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^x = e^3$

Answer: The limit is $e^3$.

Approach: Use equivalent infinitesimal substitution.

Steps:

1. As $x \to 0$, $e^x - 1 \sim x$ and $\sin x \sim x$

2. $\lim_{x \to 0} \frac{e^x - 1}{\sin x} = \lim_{x \to 0} \frac{x}{x} = 1$

Answer: The limit is 1.

Approach: Use the logarithmic limit.

Steps:

1. As $x \to 0$, $\ln(1 + x^2) \sim x^2$

2. $\lim_{x \to 0} \frac{\ln(1 + x^2)}{x^2} = 1$

Answer: The limit is 1.

Approach: Use the power limit formula.

Steps:

1. $\lim_{x \to 0} \frac{(1 + x)^3 - 1}{x} = 3$

Answer: The limit is 3.

Approach: Use the tangent limit and a change of variable.

Steps:

1. $\lim_{x \to 0} \frac{\tan 2x}{x} = \lim_{x \to 0} 2 \cdot \frac{\tan 2x}{2x} = 2 \cdot 1 = 2$

Answer: The limit is 2.

Approach: Use the cosine limit.

Steps:

1. $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

Answer: The limit is $\frac{1}{2}$.

Approach: Use a change of variable and the second important limit.

Steps:

1. Let $t = \frac{1}{x}$; then as $x \to 0$, $t \to \infty$

2. $\lim_{x \to 0} (1 + 2x)^{\frac{1}{x}} = \lim_{t \to \infty} \left(1 + \frac{2}{t}\right)^t = e^2$

Answer: The limit is $e^2$.

Approach: Use the equivalent infinitesimal of a composite function.

Steps:

1. As $x \to 0$, $e^x - 1 \sim x$

2. Hence $\sin(e^x - 1) \sim e^x - 1 \sim x$

3. $\lim_{x \to 0} \frac{\sin(e^x - 1)}{x} = \lim_{x \to 0} \frac{x}{x} = 1$

Answer: The limit is 1.

Approach: Use a change of variable and the second important limit.

Steps:

1. Let $t = x + 1$; then as $x \to \infty$, $t \to \infty$

2. $\lim_{x \to \infty} \left(1 + \frac{1}{x+1}\right)^x = \lim_{t \to \infty} \left(1 + \frac{1}{t}\right)^{t-1}$

3. $= \lim_{t \to \infty} \left(1 + \frac{1}{t}\right)^t \cdot \left(1 + \frac{1}{t}\right)^{-1} = e \cdot 1 = e$

Answer: The limit is $e$.

Approach: Use the equivalent infinitesimal of a composite function.

Steps:

1. As $x \to 0$, $\sin x \sim x$

2. Hence $e^{\sin x} - 1 \sim \sin x \sim x$

3. $\lim_{x \to 0} \frac{e^{\sin x} - 1}{x} = \lim_{x \to 0} \frac{x}{x} = 1$

Answer: The limit is 1.

Approach: Use the arcsine limit.

Steps:

1. The arcsine limit states that $\lim_{x \to 0} \frac{\arcsin x}{x} = 1$.
2. Let $t = \arcsin x$, so $x = \sin t$ and $t \to 0$ as $x \to 0$; then $\frac{\arcsin x}{x} = \frac{t}{\sin t}$.
3. The sine limit $\lim_{t \to 0} \frac{\sin t}{t} = 1$ is the reciprocal of that expression, so the ratio tends to $1$.
4. Substituting back gives $\lim_{x \to 0} \frac{\arcsin x}{x} = 1$.

Answer: The limit is 1.

Approach: Use the arctangent limit.

Steps:

1. $\lim_{x \to 0} \frac{\arctan x}{x} = 1$

Answer: The limit is 1.

Approach: Use a change of variable and the second important limit.

Steps:

1. Let $t = x^2$; then as $x \to \infty$, $t \to \infty$

2. $\lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^{x^2} = \lim_{t \to \infty} \left(1 + \frac{1}{t}\right)^t = e$

Answer: The limit is $e$.

Approach: Use a change of variable and the first important limit.

Steps:

1. Let $t = x^3$; then as $x \to 0$, $t \to 0$

2. $\lim_{x \to 0} \frac{\sin x^3}{x^3} = \lim_{t \to 0} \frac{\sin t}{t} = 1$

Answer: The limit is 1.