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Mathematical Problems and Solutions

Problem 1: Sequence Convergence

Problem Statement: Consider the sequence \( \{a_n\} \) defined by:

\[ a_n = \frac{2n^2 + 3n + 1}{n^2 + n} \]

Determine if the sequence converges, and if so, find its limit.

Solution:

To determine the limit, divide the numerator and denominator by \( n^2 \):

\[ \lim_{{n \to \infty}} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{1 + \frac{1}{n}} \]

As \( n \to \infty \), terms with \( \frac{1}{n} \) vanish:

\[ \lim_{{n \to \infty}} \frac{2 + 0 + 0}{1 + 0} = 2 \]

Thus, the sequence converges, and the limit is:

\( \boxed{2} \)

Problem 2: Function Continuity

Problem Statement: Let \( f(x) \) be a piecewise function defined as:

\[ f(x) = \begin{cases} 3x + 2 & \text{if } x \leq 1, \\ x^2 - 1 & \text{if } x > 1. \end{cases} \]

Determine if \( f(x) \) is continuous at \( x = 1 \).

Solution:

Check the left-hand and right-hand limits:

\[ \lim_{{x \to 1^-}} f(x) = 3(1) + 2 = 5, \quad \lim_{{x \to 1^+}} f(x) = (1)^2 - 1 = 0 \]

Since these limits are not equal, \( f(x) \) is not continuous at \( x = 1 \).

Problem 3: Tangent Line

Problem Statement: Find the equation of the tangent line to \( f(x) = x^3 - 3x^2 + 2x + 1 \) at \( x = 1 \).

Solution:

Compute the derivative \( f'(x) \):

\[ f'(x) = 3x^2 - 6x + 2 \]

At \( x = 1 \):

\[ f'(1) = 3(1)^2 - 6(1) + 2 = -1, \quad f(1) = 1 \]

Using point-slope form:

\[ y - 1 = -1(x - 1) \implies y = -x + 2 \]

The tangent line is:

\( \boxed{y = -x + 2} \)