Understanding Infinite Limits and Unbounded Function Behavior

As x approaches 4 from the left (x→4⁻), the expression (4-x) approaches 0 from the positive side because x is less than 4. When we divide 7 by a very small positive number, the result becomes very large and positive.

As x approaches -2 from the right (x→-2⁺), x is greater than -2, so (x+2) is positive and approaches 0. When we cube a positive number, the result is positive. When we divide 1 by a very small positive number cubed, the result becomes very large and positive.

A vertical asymptote occurs where the denominator equals zero and the numerator is not zero at that point. For f(x) = (x-2)/(x+2), the denominator equals zero when x = -2, and the numerator equals -4 at x = -2, which is not zero. Therefore, there is a vertical asymptote at x = -2.

As x approaches 0 from the left (x→0⁻), x³ approaches 0 from the negative side because cubing a negative number gives a negative result. When we divide 1 by a very small negative number, the result becomes very large and negative.

As x approaches 3 from the left (x→3⁻), x is less than 3, so (x²-9) approaches 0 from the negative side because (3)²-9 = 0 and for x slightly less than 3, x² is slightly less than 9. When we divide 2 by a very small negative number, the result becomes very large and negative.

As x approaches π/2 from the right (x→π/2⁺), the denominator (x-π/2) approaches 0 from the positive side. When we have 1 divided by a very small positive number, the result becomes very large and positive.

A vertical asymptote represents behavior where the function grows without bound as x approaches a certain value. A hole occurs when there is a factor that cancels in the numerator and denominator, resulting in a point where the function is undefined but the limit exists and is finite.

As x approaches -3 from the right (x→-3⁺), the denominator (x+3)² approaches 0 from the positive side (since squaring always gives a positive result). When we divide 4 by a very small positive number, the result becomes very large and positive.

As x approaches 5 from the left (x→5⁻), x is less than 5, so (5-x) is positive and approaches 0. When we divide 1 by a very small positive number, the result becomes very large and positive.

For a vertical asymptote at x = 2, we need the function to be undefined at x = 2 and approach infinity as x approaches 2. The expression 1/(x-2) satisfies this: it is undefined at x = 2 and approaches positive or negative infinity as x approaches 2 from either side.

As x approaches 6 from the right (x→6⁺), the denominator (x-6) approaches 0 from the positive side. When we have -2 divided by a very small positive number, the result becomes very large and negative.

When we write lim_{x→a} f(x) = ∞, we mean that as x gets arbitrarily close to a, the values of f(x) become arbitrarily large. This describes unbounded behavior, not that the function actually equals infinity (which is not a number).

As x approaches 1 from the left (x→1⁻), the denominator (1-x)² approaches 0 from the positive side (since squaring always gives a positive result). When we divide 3 by a very small positive number, the result becomes very large and positive.

To find vertical asymptotes, we set the denominator equal to zero: x²-4x+3 = 0. Factoring gives (x-1)(x-3) = 0, so x = 1 or x = 3. Both values make the denominator zero while the numerator remains 1, creating two vertical asymptotes.

As x approaches 0 from the right (x→0⁺), the denominator x approaches 0 from the positive side. When we have -1 divided by a very small positive number, the result becomes very large and negative.