Limit Exercises

Solution approach: Use the important limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

$\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x}$, as $x \to 0$, $\frac{\sin 3x}{3x} \to 1$.

Answer: The limit value is $3$.

Solution approach: The definition of an infinitesimal sequence is $\lim\limits_{n \to \infty} x_n = 0$.

$\lim\limits_{n \to \infty} \frac{1}{n} = 0$.

Answer: It is an infinitesimal sequence.

Solution approach: $1 - \cos x \sim \frac{x^2}{2}$, so the limit $\approx \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2}$.

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

Solution approach: Use the important limit $\lim\limits_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a$.

Here $a = 2$, so the limit is $e^2$.

Answer: The limit value is $e^2$.

Solution approach: As $x \to 0^+$, $f(x) = \frac{1}{x} \to +\infty$, which is an infinite quantity.

Answer: The limit does not exist (approaches $+\infty$), it is an infinite quantity.

Solution approach: First simplify, then evaluate the limit.

Detailed steps:

1. $\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2$ (when $x \neq 2$)

2. $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4$

Answer: The limit value is 4.

Solution approach: Use the addition rule and important limit.

Detailed steps:

1. $\lim_{x \to 0} \frac{\sin x + x}{x} = \lim_{x \to 0} \left(\frac{\sin x}{x} + \frac{x}{x}\right)$

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

3. $= 1 + 1 = 2$

Answer: The limit value is 2.

Solution approach: Divide numerator and denominator by the highest power of x.

Detailed steps:

1. $\frac{x^2 + 3x + 1}{x^2 + 2x} = \frac{1 + \frac{3}{x} + \frac{1}{x^2}}{1 + \frac{2}{x}}$

2. $\lim_{x \to \infty} \frac{x^2 + 3x + 1}{x^2 + 2x} = \lim_{x \to \infty} \frac{1 + \frac{3}{x} + \frac{1}{x^2}}{1 + \frac{2}{x}}$

3. $= \frac{1 + 0 + 0}{1 + 0} = 1$

Answer: The limit value is 1.

Solution approach: Calculate $\lim_{x \to 0} \frac{x^3}{x^2}$ to determine the relationship.

Detailed steps:

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

2. Since the limit is 0, $x^3$ is a higher-order infinitesimal than $x^2$.

Answer: $x^3$ is a higher-order infinitesimal than $x^2$.

Solution approach: Use equivalent infinitesimal substitution to simplify the calculation.

Detailed steps:

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

2. However, $\tan x - \sin x$ cannot be directly substituted and needs further processing

3. $\tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \sin x \left(\frac{1}{\cos x} - 1\right) = \sin x \cdot \frac{1 - \cos x}{\cos x}$

4. As $x \to 0$, $\sin x \sim x$, $1 - \cos x \sim \frac{x^2}{2}$, $\cos x \to 1$

5. So $\tan x - \sin x \sim x \cdot \frac{x^2}{2} = \frac{x^3}{2}$

6. Therefore $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{2}}{x^3} = \frac{1}{2}$

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

Solution approach: Use the boundedness of the sine function to construct a squeeze inequality.

Detailed steps:

1. Since $-1 \leq \sin \frac{1}{x} \leq 1$, we have $-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$

2. $\lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0$

3. By the squeeze theorem, $\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0$

Answer: The limit value is 0.

Solution approach: First prove that the sequence is monotonically decreasing and bounded below, then find the limit.

Detailed steps:

1. Prove monotonic decrease: $x_{n+1} - x_n = \frac{(n+1)^2 + 1}{(n+1)^2 + (n+1)} - \frac{n^2 + 1}{n^2 + n} < 0$

2. Prove bounded below: $x_n = \frac{n^2 + 1}{n^2 + n} = 1 - \frac{n-1}{n^2 + n} > 0$

3. By the monotone convergence theorem, the sequence converges

4. Find the limit: $\lim_{n \to \infty} \frac{n^2 + 1}{n^2 + n} = \lim_{n \to \infty} \frac{1 + \frac{1}{n^2}}{1 + \frac{1}{n}} = 1$

Answer: The sequence converges with limit 1.

Solution approach: Use the generalized form of the first important limit.

Detailed 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 value is 5.

Solution approach: Use the generalized form of the second important limit.

Detailed steps:

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

Answer: The limit value is $e^3$.

Solution approach: Use equivalent infinitesimal substitution.

Detailed steps:

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

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

Answer: The limit value is 1.

Solution approach: Compute the left-hand and right-hand limits separately to see if they are equal.

Detailed steps:

1. Right-hand limit: $\lim_{x \to 1^+} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^+} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1^+} (x+1) = 2$

2. Left-hand limit: $\lim_{x \to 1^-} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^-} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1^-} (x+1) = 2$

3. Since the left-hand and right-hand limits are equal, the limit exists.

Answer: The limit exists with value 2.

Solution approach: Use the definition of a limit to prove that for any $\varepsilon > 0$, there exists $N$ such that when $n > N$, $|x_n - 1| < \varepsilon$.

Detailed steps:

1. $|x_n - 1| = \left|\frac{n}{n+1} - 1\right| = \left|\frac{n-(n+1)}{n+1}\right| = \frac{1}{n+1}$

2. To make $\frac{1}{n+1} < \varepsilon$, we need $n+1 > \frac{1}{\varepsilon}$, i.e., $n > \frac{1}{\varepsilon} - 1$

3. Let $N = \left\lfloor \frac{1}{\varepsilon} - 1 \right\rfloor + 1$, then when $n > N$, $|x_n - 1| < \varepsilon$

Answer: The sequence limit is 1.

Solution approach: Compute the left-hand and right-hand limits separately to see if they are equal.

Detailed steps:

1. Right-hand limit: $\lim_{x \to 0^+} \frac{1}{x} = +\infty$

2. Left-hand limit: $\lim_{x \to 0^-} \frac{1}{x} = -\infty$

3. Since the left-hand and right-hand limits are different, the limit does not exist.

Answer: The limit does not exist.