Advanced Infinite Limits with Trigonometric, Exponential, and Logarithmic Functions

When x approaches -2, the denominator (x+2)² becomes very small. Since we're squaring the difference, the denominator is always positive regardless of whether x approaches -2 from the left or right. As the denominator approaches zero while remaining positive, the fraction 3/(x+2)² grows without bound. The constant numerator 3 doesn't change this behavior. Therefore, the limit is infinity.

When x approaches 4 from the left, x is slightly less than 4. This means x-4 is a small negative number. As x gets closer to 4 from the left, x-4 becomes smaller and smaller negative values. Dividing 2 by increasingly small negative numbers produces increasingly large negative values. Therefore, as x approaches 4 from the left, f(x) approaches negative infinity.

To evaluate this limit at infinity, we divide both numerator and denominator by the highest power of x in the denominator, which is x³. This gives us: lim(x→∞) (4 + 2/x - 5/x³)/(2 - 1/x² + 3/x³). As x approaches infinity, terms with x in the denominator (2/x, 5/x³, 1/x², and 3/x³) all approach 0. This simplifies our expression to (4 + 0 - 0)/(2 - 0 + 0) = 4/2 = 2. Therefore, the limit equals 2.

This is the definition of a vertical asymptote. If the function values grow without bound (approaching positive or negative infinity) as x gets arbitrarily close to a specific number a, then the vertical line x = a is a vertical asymptote for the graph of the function.

The tangent function is defined as tan(x) = sin(x)/cos(x). As x approaches π/2 from the right, sin(x) approaches 1 and cos(x) approaches 0 from the negative side (since cosine is negative in the second quadrant). When we divide a value approaching 1 by values approaching 0 from the negative side, the result grows without bound in the negative direction. Therefore, lim(x→(π/2)⁺) tan(x) = -∞.

Since we're approaching x = -3 from the left (x

As x approaches 0 from the left, x is negative and getting closer to zero. When we cube a negative number, the result remains negative. So x³ is a small negative number. When we divide 1 by increasingly small negative numbers, the result grows without bound in the negative direction. Therefore, lim(x→0⁻) 1/x³ = -∞.

To evaluate this limit at infinity, we compare the degrees of the polynomials in the numerator and denominator. The numerator has degree 4 (highest power is x^4) and the denominator has degree 3 (highest power is x³). When the degree of the numerator is greater than the degree of the denominator, the limit at infinity will be infinite. Since the leading coefficient of the numerator (1) is positive, and for large positive x values, the function will be positive, the limit approaches positive infinity.

First, we factor both numerator and denominator. The numerator x³ - 1 = (x - 1)(x² + x + 1) using the difference of cubes formula. The denominator x² - 2x + 1 = (x - 1)². This gives us: lim(x→1) (x - 1)(x² + x + 1)/(x - 1)² = lim(x→1) (x² + x + 1)/(x - 1). As x approaches 1, the numerator approaches 1² + 1 + 1 = 3, while the denominator approaches 0. Since the denominator approaches 0 while the numerator approaches a non-zero value, the limit is infinite. Specifically, as x approaches 1 from the right, the denominator is positive, so the limit is +∞.

As x approaches 3 from the left, the denominator x² - 9 approaches 0 from the negative side because x² will be slightly less than 9. The numerator 2x² - 3x + 1 approaches 2(9) - 3(3) + 1 = 18 - 9 + 1 = 10, which is positive. When we divide a positive number (approximately 10) by increasingly small negative numbers, the result grows without bound in the negative direction. Therefore, as x approaches 3 from the left, f(x) approaches negative infinity.

To evaluate this limit at negative infinity, we divide both numerator and denominator by the highest power of x, which is x⁵. This gives us: lim(x→-∞) (4 + 2/x² - 1/x^4)/(2 - 3/x³ + 1/x⁵). As x approaches negative infinity, all terms with x in the denominator (2/x², 1/x^4, 3/x³, and 1/x⁵) approach 0. This simplifies our expression to (4 + 0 - 0)/(2 - 0 + 0) = 4/2 = 2. Therefore, the limit equals 2.

As x approaches 1 from the right, both the numerator ln(x) and the denominator (x-1) approach 0, giving us the indeterminate form 0/0. We can apply L'Hôpital's Rule by taking derivatives of numerator and denominator: the derivative of ln(x) is 1/x, and the derivative of (x-1) is 1. This gives us lim(x→1⁺) (1/x)/1 = lim(x→1⁺) 1/x = 1/1 = 1. Therefore, as x approaches 1 from the right, f(x) approaches 1.

As x approaches 0, both the numerator (1 - cos(x)) and the denominator x² approach 0, giving us the indeterminate form 0/0. We can apply L'Hôpital's Rule by taking derivatives of numerator and denominator: the derivative of (1 - cos(x)) is sin(x), and the derivative of x² is 2x. This gives us lim(x→0) sin(x)/(2x). This is still 0/0, so we apply L'Hôpital's Rule again: the derivative of sin(x) is cos(x), and the derivative of 2x is 2. This gives us lim(x→0) cos(x)/2 = cos(0)/2 = 1/2. Therefore, the limit equals 1/2.

This limit is in the indeterminate form ∞/∞, so we can apply L'Hôpital's Rule. Taking derivatives of numerator and denominator gives us lim(x→∞) (1/x)/(1/(2√x)) = lim(x→∞) (1/x)·(2√x/1) = lim(x→∞) 2/√x. As x approaches infinity, √x also approaches infinity, so 2/√x approaches 0. Therefore, the original limit equals 0.

First, we factor both numerator and denominator. The numerator x² - 4 = (x - 2)(x + 2). The denominator x³ - 8 = (x - 2)(x² + 2x + 4) using the difference of cubes formula. This gives us f(x) = (x - 2)(x + 2)/[(x - 2)(x² + 2x + 4)]. For x ≠ 2, we can cancel the (x - 2) factors, resulting in f(x) = (x + 2)/(x² + 2x + 4). At x = 2, both numerator and denominator of the original function are zero, but after simplification, the function approaches (2 + 2)/(4 + 4 + 4) = 4/12 = 1/3. This implies the graph has a hole at x = 2, not a vertical asymptote. The quadratic x² + 2x + 4 in the denominator has discriminant b² - 4ac = 4 - 16 = -12, which is negative, so it has no real roots. Therefore, the function has no vertical asymptotes.