Possible results

1. Finite Limit (number): If the limit of the function exists and is a finite number, it means that as the input values get closer and closer to a certain point, the function values approach a specific finite value. The function is approaching the same value from the left and the right.

2. Infinite Limit (-∞ or ∞): If the limit of the function becomes infinite (positive or negative) as the input approaches a particular value, it means that the function values grow without a bound as the input values get closer to that point (when the function has vertical asymptotes or unbounded).

3. Limit Does Not Exist: The limit of a function may not exist at a particular point if the function's values do not approach a single value as the input gets arbitrarily close to that point. This can happen when the function oscillates or behaves erratically around the point.

4. Approaching from Different Directions: Sometimes, the limit might be different depending on whether you approach the point from the left or the right side. In this case, the function might have a "jump" or a discontinuity at that point.

5. Approaching Different Values: The limit can also vary for different sequences or paths of approaching the point. This can indicate that the function has a more complex behavior near that point, possibly involving oscillations or fractal-like patterns.

6. Indeterminate Form: In certain cases, you might encounter an indeterminate form (like 0/0 or ∞/∞) when evaluating the limit. This means that more analysis is needed to determine the actual behavior of the function at that point. Techniques like L'Hôpital's Rule can help in simplifying and evaluating these indeterminate forms.

7. Limit at Infinity: Evaluating the limit as the input approaches infinity or negative infinity can provide insights into the long-term behavior of the function. Depending on whether the limit is finite, infinite, or non-existent, you can determine how the function grows or decays as the input becomes very large or very small.

   1. $$\lim_{x\to\ \infty} f(x)$$

   2. $$\lim_{x\to\ -\infty} f(x)$$

8. Special Cases: Occasionally, you might encounter specific cases where the function has a unique limit that requires careful analysis, such as limits involving trigonometric, exponential, or logarithmic functions.

Why it is important

Limits allow us to understand and analyze the behavior of functions as they approach specific values or as their inputs get arbitrarily close to certain points.

It is important to understand the idea of continuity.

A function is continuous at a point if its limit at that point exists and equals its value at that point. This concept is essential in analyzing the smoothness and connectedness of functions.

The derivative of a function at a point is defined using a limit. Many functions are not defined in simple algebraic terms, but their derivatives can still be calculated using limits.

Limits help us understand rates of change in various contexts, such as physics, economics, and engineering. Many real-world phenomena involve quantities that change continuously, such as populations, temperatures, and stock prices. Limits allow us to model and analyze these changes more accurately.

Evaluating Limits Methods

• Substitution Method: simply plug in the number x is approaching to

$$\lim_{x\to\ -1} \frac{1}{x}=\frac{1}{-1}=-1 \, \to\, limit\,\, exists$$

$$\lim_{x\to\ 0} \frac{1}{x}=undefined \, \to\, limit\,\, does\,\, not\,\, exist$$

$$because\, \lim_{x\to\ 0^-} \frac{1}{x}=-\infty \,\, but \,\, \lim_{x\to\ 0^+} \frac{1}{x}=\infty$$

if you get a number answer, the limit exists

if undefined, then use another method

if the limit from the left does not match the limit from the right, then the limit does not exist

• Factoring Method: factor out, then cancel to get rid of the problem area

$$\lim_{x\to\ 0} \frac{4x^5-x^2}{x^2}=3^{-1}=\lim_{x\to\ 0}\frac{x^2(4x^3-1)}{x^2}$$

$$=\lim_{x\to\ 0} 4x^3-1=4*0^3-1=-1 \,\to\,limit \,\,exists$$

• Conjugate Method: look for the same parts but with different signs to get rid of the problem (usually involves square root)

$$\lim_{x\to\ 16} \frac{x-16}{\sqrt{x}-4}=\lim_{x\to\ 16}\frac{x-16}{\sqrt{x}-4} * \frac{\sqrt{x}+4}{\sqrt{x}+4}=$$

$$=\lim_{x\to\ 16}\frac{(x-16)(\sqrt{x}+4)}{(\sqrt{x})^2-4^2}=\lim_{x\to\ 16}\sqrt{x}+4=$$

$$=\sqrt{16}+4=8 \to limit \, \, exists$$

Problems

1. Try it yourself (the answer is at the end of the post):

$$\lim_{x\to\ e} ln(x)+1$$

1. Try it yourself (the answer is at the end of the post):

$$\lim_{x\to\ 1}(\frac{7x^2-10+3}{x-1}+\frac{2-7x}{x+1})$$

1. Try it:

$$\lim_{x\to\ 9} \frac{3}{\sqrt{x}+3}$$

In the end

1. Substitution Method

$$\lim_{x\to\ e} ln(x)+1=ln(e)+1=2$$

1. Separate the limit into two different limits, the first one is the Factoring Method and the second is the Substitution Method right away

$$\lim_{x\to\ 1}(\frac{7x^2-10+3}{x-1}+\frac{2-7x}{x+1})=\lim_{x\to\ 1}\frac{7x^2-10+3}{x-1}+\lim_{x\to\ 1}\frac{2-7x}{x+1}$$

$$\lim_{x\to\ 1} \frac{(7x-3)(x-1)}{x-1}+(-\frac{5}{2})=7(1)-3-\frac{5}{2}=\frac{3}{2}$$

1. Substitution Method (if you tried to go the Conjugate route b/c you saw a square root, remember always try substitution first!)

$$\lim_{x\to\ 9} \frac{3}{\sqrt{x}+3}= \frac{3}{\sqrt{9}+3}=\frac{3}{3+3}=\frac{1}{2}$$