Infinite Limits with Rational Functions and Vertical Asymptotes

As x approaches 7 from the left (x→7⁻), the expression (7-x) approaches 0 from the positive side because x is less than 7. When we divide 5 by a very small positive number, the result becomes very large and positive.

As x approaches -4 from the right (x→-4⁺), the denominator (x+4)² approaches 0 from the positive side (since squaring always gives a positive result). When we divide 1 by a very small positive number, the result becomes very large and positive.

For f(x) = 1/(x²-4), the denominator equals zero when x²-4 = 0, which gives x = 2 or x = -2. Both values create vertical asymptotes because the numerator is 1 (not zero) at these points.

As x approaches 0 from the left (x→0⁻), x² approaches 0 from the positive side because squaring a negative number gives a positive result. When we divide 1 by a very small positive number, the result becomes very large and positive.

As x approaches 2 from either side, the denominator (x-2)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, regardless of whether we approach from the left or right.

As x approaches 3 from the right (x→3⁺), x is greater than 3, so (3-x) is negative and approaches 0. When we divide 1 by a very small negative number, the result becomes very large and negative.

As x approaches 2 from the right (x→2⁺), the expression (2-x) approaches 0 from the negative side because x is greater than 2. When we cube a negative number, the result is negative. When we divide 6 by a very small negative number, the result becomes very large and negative.

To find vertical asymptotes, we set the denominator equal to zero: x³-8x = 0. Factoring gives x(x²-8) = 0, which gives x = 0, x = 2√2, or x = -2√2. All three values make the denominator zero while the numerator remains 1, creating three vertical asymptotes.

As x approaches 0 from the right (x→0⁺), the denominator x³ approaches 0 from the positive side (since cubing a positive number gives a positive result). When we divide 5 by a very small positive number, the result becomes very large and positive.

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

A function has a vertical asymptote at x = a if (1) the function is undefined at x = a (typically because the denominator is zero), (2) the numerator is not also zero at x = a (to distinguish from holes), and (3) the function approaches positive or negative infinity as x approaches a.

As x approaches 8 from the right (x→8⁺), the denominator (x-8) approaches 0 from the positive side. When we have -3 divided by a very small positive number, the result becomes very large and negative.

For f(x) = 1/x², as x approaches 0 from either the left or right, x² approaches 0 from the positive side (since squaring gives positive results). Therefore, 1/x² approaches positive infinity from both sides. For the other functions, the behavior differs between left and right approaches.

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 in the negative direction. This describes unbounded decreasing behavior, not that the function actually equals negative infinity.

As x approaches -5 from the left (x→-5⁻), x is less than -5, so (x+5) is negative and approaches 0. When we divide 2 by a very small negative number, the result becomes a very large negative number.