# Functions (mathematics)/Limit
## Definition

A mathematical limit denotes the output of a function f(x) as it approaches (gets infinitely close to) some input x. The notation used to express a limit is $\lim _{x\to a}f(x)=b$ and is read as "the limit as x approaches a of f(x) is equal to b". This is called a two-sided limit because it tells us that the function approaches the same value b when the input approaches a from the left (less than a) and to the right (greater than a). However, one-sided limits also exist and are important in the evaluation of two-sided limits. To express that the limit as x approaches a from the left of f(x) is equal to some value b, we write $\lim _{x\to a^{-}}f(x)=b$. Note the negative sign by a to indicate that this is a one-sided limit as we approach the input a from the left (with inputs less than a). Similarly, $\lim _{x\to a^{+}}f(x)=b$ indicates that the function approaches the output b as we approach from the right side with input values greater than a. If both one-sided limits approaching from both sides of some value a are equal to the same value b, then the two-sided limit exists and equals b. Otherwise, the two-sided limit does not exist (DNE). That is, if $\lim _{x\to a^{-}}f(x)=b$ and $\lim _{x\to a}f(x)=c$ with $b\neq c$, then $\lim _{x\to a}f(x)=DNE$. Limits are often used in definitions for mathematical concepts in calculus as we will see in the following chapters with the derivative and antiderivative (integral).

## Evaluating Limits
### Direct Substitution

Sometimes, especially for continuous functions, simply evaluating the function at our desired input will give us the correct result. For example, to evaluate $\lim _{x\to 3}{2x^{2}+1}$, we can just plug in 3 for x to get $\lim _{x\to 3}{2x^{2}+1}=2(3)^{2}+1=19$. This method should be the first one you try when asked to evaluate a limit.

### Factoring

If substituting fails, look to see if part of the function can be factored. By factoring the numerator or denominator of a function, we are often times able to eliminate a factor that is causing our output to be undefined, and can then successfully apply direct substitution. For example, attempting to evaluate $\lim _{x\to 5}{\frac {x^{2}-25}{x-5}}$ using substitution gives us ${\frac {0}{0}}$ which is undefined. However, if we factor $x^{2}-25$ into $(x+5)(x-5)$, the two $x-5$ terms cancel out and we get $\lim _{x\to 5}{x+5}$ which can easily be solved using direct substitution. This method is useful when the function has a "hole" at the value we are evaluating the limit at. This means the one-sided limits from either side of the value are equal, but the output at that point is undefined. By factoring and cancelling out terms, we effectively "plug" that hole and produce a function that is continuous at that point.

### Multiplying by the Conjugate

Another strategy to have in your toolbox when direct substitution fails is multiplying a function by the conjugate. When the function is fractional and contains a square root term, this method is often helpful. A conjugate is simply a two-term expression (binomial) in which the sign operator combining the terms is flipped. For example, the conjugate of $2x^{3}+x$ is $2x^{3}-x$. The following limit cannot be evaluated via substitution:
$\lim _{x\to 9}{\frac {3-{\sqrt {x}}}{x-9}}$
However, we recognize that the numerator contains a binomial expression with a square root, so it may be a good idea to attempt to multiply the function by the conjugate of the numerator, $3+{\sqrt {x}}$. To avoid changing the function (and thus the limit), we need to multiply both the numerator and denominator by this conjugate. This is equivalent to multiplying the function by 1 and will not change the identity of the function. Doing so gives us the following:
$\lim _{x\to 9}{{\frac {(3-{\sqrt {x}})}{(9-x)}}\cdot {\frac {(3+{\sqrt {x}})}{(3+{\sqrt {x}})}}}=\lim _{x\to 9}{\frac {9-x}{(9-x)(3+{\sqrt {x}})}}=\lim _{x\to 9}{\frac {1}{3+{\sqrt {x}}}}$
Notice how by multiplying the numerator by its conjugate, the square root term is eliminated and allows us to cancel out the resulting $9-x$ term. This gives us a limit that we can now solve with direct substitution:
$\lim _{x\to 9}{\frac {1}{3+{\sqrt {x}}}}={\frac {1}{3+{\sqrt {9}}}}={\frac {1}{6}}$

## Limits at Infinity

Often times it is of interest to look at a function's output as the input becomes infinitely large or infinitely small. This can help us understand the behavior of the function in the long-term (with increasingly positive/negative inputs) and can tell us if the function increases/decreases without bound, or if it approaches some finite value. Frequently, we are interested in the behavior of rational functions at infinity. Rational functions are written as dividing one polynomial function by another. For example,
$f(x)={\frac {3x-1}{x^{2}+1}}$
is a rational function. To determine the limit at infinity of a rational function, we start by looking at the degrees of the two polynomials in the numerator and denominator. Remember from algebra that the degree of a polynomial is simply the highest power a term in the polynomial is raised to. In the example above, we see that the numerator $3x-1$ is a degree-1 polynomial, while the denominator $x^{2}+1$ is degree 2. Since the degree of the denominator is greater than that of the numerator, we call this a "bottom-heavy" function. This means that the denominator, or bottom part of our function, will grow significantly faster than the top as our input increases to infinity. Intuitively, with very large value of x, we can see that this will result in our function trending towards 0 and so
$\lim _{x\to \infty }{\frac {3x-1}{x^{2}+1}}=0$
We've now seen what happens when the function is bottom-heavy, but what about when the numerator's degree is greater than the denominator's? Or when the degrees' are equal? Well, when the degree of the numerator is greater than that of the denominator, we call the function "top-heavy" since the numerator will outpace (grow faster) than the denominator for increasingly large inputs of x. In this case, the limit will equal infinity or DNE (think about why this is). In the case that the degree's are equal, the function is called "equal-heavy" and the numerator and denominator grow at roughly equal pace and thus the limit will approach a finite value. This value is simply the ratio of the coefficients of the highest term in the numerator's polynomial and denominator's polynomial. For example, consider the following limit:
$\lim _{x\to \infty }{\frac {6x^{3}+2x-5}{2x^{3}-1}}$
First, we notice that the degree of the numerator and denominator are both equal to 3. Now, we can take the coefficients of the highest term from both the top and bottom to determine the precise value of our limit. This gives us:
$\lim _{x\to \infty }{\frac {6x^{3}+2x-5}{2x^{3}-1}}={\frac {6}{2}}=3$
While this was a simple example, infinite limits of equal-heavy limits may require you to do a little extra work before you can determine the leading coefficients. For example, the following limit may not appear to even be equal-heavy at first glance:
$\lim _{x\to \infty }{\frac {\sqrt {3x^{2}+x}}{4x+3}}$
However, remember that we only care about the highest powered terms in the numerator and denominator. As such, we can ignore the $x$ term under the square root in the numerator and everything but the $x^{2}$ term in the denominator. This allows us to rewrite the limit as:
$\lim _{x\to \infty }{\frac {{\sqrt {3}}x}{4x}}$
which we can easily see is the limit of an equal-heavy function approaching ${\frac {\sqrt {3}}{4}}$

### Other Limits at Infinity

So far, we have only learned how to find the limit as our input approaches infinity of rational functions. However, we can also determine the limit at infinity of non-rational functions.

## Continuity at a Point

There exists a three-step definition of continuity at a point involving limits. If the following three conditions are true, then the function $f(x)$ is continuous at $x=a$:
* $f(a)$ exists
* $\lim _{x\to a}f(x)$ exists
* $\lim _{x\to a}f(x)=f(a)$