Limits and Approaching Values

Approaching Two from the Right

• Consider $x$ approaching 2 from the right, so x > 2.

• Create a table with values of $x$ slightly greater than 2 and calculate $f(x) = x^2 - 1$.

• Example table:

  • $x = 2.1$, $f(x) = (2.1)^2 - 1 = 3.41$

  • $x = 2.01$, $f(x) = (2.01)^2 - 1 = 3.0401$

  • $x = 2.001$, $f(x) = (2.001)^2 - 1 = 3.004$

  • $x = 2.0001$, $f(x) = (2.0001)^2 - 1 = 3.0004$

  • $x = 2.00001$, $f(x) = (2.00001)^2 - 1 = 3.00004$

• As $x$ gets closer to 2 from the right, $f(x)$ approaches 3.

Behavior of f(x) as x Approaches 2

• As $x$ approaches 2 from the left, $f(x)$ approaches 3.

• As $x$ approaches 2 from the right, $f(x)$ approaches 3.

• Since the behavior of $f(x)$ is the same from both sides, we conclude that as $x$ approaches 2, $f(x)$ approaches 3.

• This is written as $\lim<em>{x \to 2} f(x) = 3$, which means $\lim</em>{x \to 2} (x^2 - 1) = 3$.

Definition of a Limit

• Suppose $f(x)$ is defined when $x$ is near $a$, meaning $f(x)$ is defined on some open interval containing $a$, except possibly at $a$ itself.

• We write $\lim_{x \to a} f(x) = L$ and say "the limit of $f(x)$ as $x$ approaches $a$ equals $L$" if we can make the values of $f(x)$ arbitrarily close to $L$ by restricting $x$ to be sufficiently close to $a$ on either side, but not equal to $a$.

• In the example, we made $x$ sufficiently close to 2 from the right (e.g., 2.1, 2.01, 2.001) and from the left.

Existence of a Limit

• If $f(x)$ approaches different numbers as $x$ approaches $a$ from the left and from the right, then the limit of $f(x)$ as $x$ approaches $a$ does not exist. The behavior of $f$ should be the same no matter whether x approaches from the left or from the right for the limit to exist.

• The function need not be defined at $x = a$, but we consider values of $x$ near $a$.

Estimating a Limit: Example

• Estimate the value of $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.

• Here, $a = 3$ and $f(x) = \frac{x^2 - 9}{x - 3}$.

Observation

• $f(3) = \frac{3^2 - 9}{3 - 3} = \frac{0}{0}$, which is undefined. So, $f(x)$ is not defined at $x = 3$.

• However, the limit may still exist.

• $f(x)$ is defined on an open interval containing 3, but not at 3 itself.

Approaching 3 from the Left (x < 3)

• Create a table:

  • $x = 2.5$, $f(x) = \frac{(2.5)^2 - 9}{2.5 - 3} = \frac{6.25-9}{-0.5} = \frac{-2.75}{-0.5} = 5.5$

  • $x = 2.9$, $f(x) = \frac{(2.9)^2 - 9}{2.9 - 3} = \frac{8.41 - 9}{-0.1} = \frac{-0.59}{-0.1} = 5.9$

  • $x = 2.99$, $f(x) = 5.99$

  • $x = 2.999$, $f(x) = 5.999$

• As $x$ approaches 3 from the left, $f(x)$ approaches 6.

Approaching 3 from the Right (x > 3)

• Create a table:

  • $x = 3.1$, $f(x) = \frac{(3.1)^2 - 9}{3.1 - 3} = \frac{9.61 - 9}{0.1} = \frac{0.61}{0.1} = 6.1$

  • $x = 3.01$, $f(x) = 6.01$

  • $x = 3.001$, $f(x) = 6.001$

  • $x = 3.0001$, $f(x) = 6.0001$

• As $x$ approaches 3 from the right, $f(x)$ approaches 6.

Conclusion

• From the tables, $f(x)$ approaches 6 as $x$ approaches 3.

• Thus, $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$.

Estimating Limits: Example 2

• Estimate $\lim_{x \to 0} \frac{\sin(x)}{x}$.

• If we attempt to directly substitute, we get an indeterminate form $\frac{0}{0}$.

• The goal is to find the estimates and limits as x approaches zero.

Approaching 0 from the Left (x < 0)

• Because we are approaching 0 from the left, x has to be negative.

• Create a table with negative values approaching zero and evaluate corresponding values:

  • x = -1, f(x) = sin(-1) / -1 ≈ 0.017

  • x = -0.5, f(x) = sin(-0.5) / -0.5 ≈ 0.0087

  • x = -0.1, f(x) = sin(-0.1) / -0.1 ≈ 0.01745328366

  • x = -0.01, f(x) = sin(-0.01) / -0.01 ≈ 0.0745

  • x = -0.001, f(x) = sin(-0.001) / -0.001 ≈ 0.017