Limits Exercise List 01

Introductory limits exercises. Focus on applying the basic properties and recognizing fundamental limits.

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Exercise 1

Calculate:

$$\lim_{x \to 3} (2x^2 - 5x + 1)$$

View solution

Since the function is a polynomial (continuous), we simply substitute the value:

$$\lim_{x \to 3} (2x^2 - 5x + 1) = 2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4$$

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Exercise 2

Calculate:

$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$

View solution

We have an indeterminate form of the type $\frac{0}{0}$. We factor the numerator:

$$\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x+2 \quad (x \neq 2)$$

Therefore:

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

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Exercise 3

Calculate:

$$\lim_{x \to 0} \frac{\sin x}{x}$$

View solution

This is a fundamental limit:

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

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Exercise 4

Calculate:

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

View solution

We use the identity $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$ or the Taylor expansion. A classical approach is:

$$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \frac{1 + \cos x}{1 + \cos x} = \lim_{x \to 0} \frac{\sin^2 x}{x^2(1 + \cos x)} = \frac{1}{2}$$

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Exercise 5

Calculate:

$$\lim_{x \to +\infty} \frac{3x + 1}{x}$$

View solution

We divide the numerator and denominator by $x$:

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

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Exercise 6

Calculate:

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

View solution

We factor the numerator and denominator:

$$\frac{x^2 - 1}{x^2 + 2x - 3} = \frac{(x-1)(x+1)}{(x-1)(x+3)} = \frac{x+1}{x+3} \quad (x \neq 1)$$

Therefore:

$$\lim_{x \to 1} \frac{x+1}{x+3} = \frac{2}{4} = \frac{1}{2}$$

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Exercise 7

Calculate the one-sided limits and the limit, if it exists:

$$\lim_{x \to 0} \frac{|x|}{x}$$

View solution

• Right-hand limit: $\displaystyle\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1$
• Left-hand limit: $\displaystyle\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1$

Since the one-sided limits are different, the limit does not exist.

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Exercise 8

Calculate:

$$\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$$

View solution

We rationalize the numerator:

$$\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{x - 4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}$$

Therefore:

$$\lim_{x \to 4} \frac{1}{\sqrt{x}+2} = \frac{1}{4}$$