Infinite Limits at Infinity for Polynomial, Rational, and Exponential Functions

We examine the behavior of the function as x approaches 3. When x is close to 3, the denominator (x-3)² becomes very small. Since we're squaring the difference, the denominator is always positive regardless of whether x approaches 3 from the left or right. As the denominator approaches zero while remaining positive, the fraction 1/(x-3)² grows without bound. Therefore, the limit is infinity.

When x approaches -2 from the right, x is slightly greater than -2. This means x+2 is a small positive number. As x gets closer to -2 from the right, x+2 becomes smaller and smaller positive values. Dividing 1 by increasingly small positive numbers produces increasingly large positive values. Therefore, as x approaches -2 from the right, f(x) approaches positive infinity.

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

The graph indicates that as x approaches 1 from the left, function values increase without bound toward positive infinity. Therefore the left-hand limit is +∞.

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

Since we're approaching x = 1 from the left (x

As x approaches 0 from the right, x is positive and getting smaller. The square root of a small positive number is also a small positive number. When we divide 1 by increasingly small positive numbers, the result grows without bound in the positive direction. Therefore, lim(x→0⁺) 1/√x = ∞.

To evaluate this limit at infinity, we compare the degrees of the polynomials in the numerator and denominator. The numerator has degree 3 (highest power is x³) and the denominator has degree 2 (highest power is x²). 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 (2) 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 x² - 4x + 4 = (x - 2)². The denominator x³ - 8 = (x - 2)(x² + 2x + 4) using the difference of cubes formula. This gives us: lim(x→2) (x - 2)²/[(x - 2)(x² + 2x + 4)] = lim(x→2) (x - 2)/(x² + 2x + 4). Now we can substitute x = 2 directly: (2 - 2)/(2² + 2(2) + 4) = 0/(4 + 4 + 4) = 0/12 = 0. Therefore, the limit equals 0.

As x approaches 2 from the right, the denominator x² - 4 approaches 0 from the positive side because x² will be slightly greater than 4. The numerator 3x² + 2x - 1 approaches 3(4) + 2(2) - 1 = 12 + 4 - 1 = 15, which is positive. When we divide a positive number (approximately 15) by increasingly small positive numbers, the result grows without bound in the positive direction. Therefore, as x approaches 2 from the right, f(x) approaches positive infinity.

To evaluate this limit at negative infinity, we divide both numerator and denominator by the highest power of x, which is x^4. This gives us: lim(x→-∞) (5 + 2/x² - 1/x^4)/(3 - 1/x + 7/x^4). As x approaches negative infinity, all terms with x in the denominator (2/x², 1/x^4, 1/x, and 7/x^4) approach 0. This simplifies our expression to (5 + 0 - 0)/(3 - 0 + 0) = 5/3. Therefore, the limit equals 5/3.

As x approaches 1 from the left, x is slightly less than 1, so x-1 is a small negative number. The numerator e^x approaches e^1 = e, which is positive. When we divide a positive number (approximately e) by increasingly small negative numbers, the result grows without bound in the negative direction. Therefore, as x approaches 1 from the left, f(x) approaches negative infinity.

For x ≠ 0, we can rewrite |x|/x² as 1/|x| because |x|/x² = |x|/(|x|·|x|) = 1/|x|. As x approaches 0 from either side, |x| approaches 0 from the positive side. When we divide 1 by increasingly small positive numbers, the result grows without bound in the positive direction. Therefore, lim(x→0) |x|/x² = ∞.

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→∞) (2x + 3)/e^x. This is still ∞/∞, so we apply L'Hôpital's Rule again: lim(x→∞) 2/e^x. As x approaches infinity, e^x grows much faster than any polynomial, so 2/e^x approaches 0. Therefore, the original limit equals 0.

First, we factor both numerator and denominator. The numerator x² - 1 = (x - 1)(x + 1). The denominator x³ - x = x(x² - 1) = x(x - 1)(x + 1). This gives us f(x) = (x - 1)(x + 1)/[x(x - 1)(x + 1)]. For x ≠ 1 and x ≠ -1, we can cancel common factors, resulting in f(x) = 1/x. At x = 1 and x = -1, both numerator and denominator are zero. However, because the factors (x-1) and (x+1) cancel out completely from the denominator, the limit exists and is finite at these points, creating holes rather than vertical asymptotes. At x = 0, the function is undefined and the limit approaches infinity, indicating a vertical asymptote. Therefore, the only vertical asymptote is at x = 0.