What Are Limits in Calculus? | Outlier (2023)

In This Article

1. What Is a Limit?

2. Limit Notation

3. Limits in Differential Calculus and Integral Calculus

4. 7 Exercises for You to Try Now (with Solutions)

What Is a Limit?

Limits are the foundation of calculus. Understanding how to do limits in calculus is crucial for understanding other fundamental concepts in calculus, such as differentiation and integration. Given a function $f$f, a limit is the value that $f(x)$f(x) approaches as $x$x approaches some value.

For example, take the function $f(x) = 2x$f(x)=2x, graphed below.

Suppose we want to find the limit of $f$f at $x = 2$x=2. We want to find the $y$y-value that $f$f approaches as $x$xgets infinitely close to 2. Looking at the graph, it’s clear that $y$yapproaches 4 as $x$x gets closer and closer to 2. In this case, we say that the limit as $x$x approaches 2 is 4.

It’s important to clarify that the value of a function at $x$x and the limit of a function at $x$x are two different concepts. In the above example, the value of $f$f at 2 equals the limit of $f$f as $x$x approaches 2. However, this is not always the case. The difference is that the limit as $x$x approaches some $c$c can often be evaluated even when $f$f is not defined at $c$c.

For example, consider the function $f(x) = \frac{x^2-1}{x-1}$f(x)=x−1x2−1​, graphed below. Notice that there is a removable discontinuity called a hole at $x = 1$x=1, since we can’t have a denominator of zero.

Even though $f(x)$f(x) is not defined at 1, we can still use limits to understand the behavior of the function as $x$x approaches 1. As $x$x moves infinitely close to $1$1, we see that $y$y moves infinitely closer to 2, yet never reaches it. It may be helpful to refer to the table of function values below to see this behavior clearly.

So, we can say that the limit of $f$f as $x$x approaches 1 is 2, even though the value of $f$f at $1$1 is undefined. Notice that the value of $f$f as $x$x approaches 1 gets arbitrarily close to 2 from both directions. This means that if we approach 1 from either the negative or positive direction, $y$y still approaches 2.

You can verify this visually by looking at the graph of $f$f illustrated earlier.

Limit Notation

If we want to evaluate the limit of a real-valued function $f$f as $x$x approaches some real number $c$c, we use the following notation:

$\lim_{x\to c}f(x) = L$limx→c​f(x)=L

In this notation, “$\lim$lim" indicates the operation of taking a limit. Underneath, “ $x\to c$x→c” represents the value that $x$x approaches, at which we want to evaluate the limit. Finally, $L$L represents the solution to the limit as $f(x)$f(x) approaches $c$c.

Limits in Differential Calculus and Integral Calculus

Why do we use limits in calculus? Limits are useful in calculus because they permit us to examine the behavior of a function around a given x-value even when the function is not defined at that x-value.

Limits also form the foundation of calculus. The principles of calculus were originally constructed on infinitesimals, which were eventually considered too imprecise to constitute the basis of calculus. Limits were created to remedy these imprecise arguments in calculus with more mathematically precise arguments. Now, limits define the most principal concepts in calculus.

For example, the definition of a derivative is a limit. A function is called differentiable at $x$x if the following limit exists:

$\mathop {\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x \right)}}{\Delta{x} } = L$Δx→0lim​Δxf(x+Δx)−f(x)​=L

If the limit $L$L exists, we say that the derivative of the function $f$f at $x$x is $L$L, and we can write $\frac{d}{{dx}}f\left( x \right) = L$dxd​f(x)=L. This limit represents the instantaneous rate of change of $f$f at $x$x.

Limits are also used to define integrals. The limit definition of the definite integral of $f$f on the interval $[a, b]$[a,b] is:

$\int_{a}^{b} f(x) \,dx = \lim_{n\to\infty}\sum_{i=1}^{n}f(c_1)\Delta{x_i} = L$∫ab​f(x)dx=limn→∞​∑i=1n​f(c1​)Δxi​=L

In this case, the limit $L$L represents the area under the curve $f(x)$f(x) on the interval between $a$a and $b$b. We won’t go into detail about this syntax here, but it’s good to recognize that limits are the foundation of the two most essential operations in calculus.

7 Exercises for You to Try Now (with Solutions)

Here are some exercises to practice and their solutions, as well as some useful properties of limits and techniques for evaluating limits.

Exercise 1

Using the graph of $f$f below, what is a good estimate for $\lim_{x\to 1}f(x)$limx→1​f(x)?

Solution: $\lim_{x\to 1}f(x) = 1$limx→1​f(x)=1. Although $f(1) = 3$f(1)=3, the graph indicates that $f(x)$f(x) approaches 1 on both sides as $x$x approaches 1. This is an example of a scenario when the limit of a function at $x$x does not equal the value of the function at $x$x.

Exercise 2

Using the graph of $f$fbelow, what is a good estimate for $\lim_{x\to 2}f(x)$limx→2​f(x)?

Solution: The limit does not exist. In the positive direction, $f(x)$f(x) approaches 5 as $x$x approaches 2. In the negative direction, $f(x)$f(x) approaches 7 as $x$x approaches 2. So, we can say that $\lim_{x\to 2^+}f(x) = 5$limx→2+​f(x)=5 and $\lim_{x\to 2^-}f(x) = 7$limx→2−​f(x)=7. However, since the limit does not approach the same value from both directions, the limit does not exist.

Exercise 3

Let $f(x) = x^2 + 4x -3$f(x)=x2+4x−3. Evaluate $\lim_{x\to 2}f(x)$limx→2​f(x).

To evaluate this limit, we can note four important properties of limits:

• $\lim_{x\to a}cf(x) = c\lim_{x\to a}f(x)$limx→a​cf(x)=climx→a​f(x)

The Constant Multiple Rule states that we can bring the constant $c$c to the outside of the limit.

• $\lim_{x\to a}f(x) \pm g(x) = \lim_{x\to a}f(x) \pm \lim_{x\to a}g(x)$limx→a​f(x)±g(x)=limx→a​f(x)±limx→a​g(x)

The Sum Rule states that the limit of the sum of two functions is equal to the sum of their limits.

• $\lim_{x\to a}x = a$limx→a​x=a

The limit of $x$x as $x$x approaches $a$a is equal to $a$a.

• $\lim_{x\to a}c = c$limx→a​c=c

The limit of a constant is equal to the constant itself, no matter what $a$ais.

Using these properties, we have:

$\lim_{x\to 2}x^2 + 4x -3 = \lim_{x\to 2}x^2 + 4\lim_{x\to 2}x - \lim_{x\to 2}3$limx→2​x2+4x−3=limx→2​x2+4limx→2​x−limx→2​3

$= (2)^2 + 4(2) - 3$=(2)2+4(2)−3

$= 9$=9

Notice that in this example, $f(2) = \lim_{x\to 2}f(x)$f(2)=limx→2​f(x). As noted before, this is not the case for every function. However, when your function is a polynomial, you can always say that $\lim_{x\to a}f(x) = f(a)$limx→a​f(x)=f(a), since polynomials are always continuous.

Exercise 4

Let $f(x) = \frac{2x+4}{x^3}$f(x)=x32x+4​. Evaluate $\lim_{x\to3}f(x)$limx→3​f(x).

For this problem, we can use the Quotient Rule, which states that $\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{ \lim_{x\to a}f(x)}{ \lim_{x\to a}g(x)}$limx→a​g(x)f(x)​=limx→a​g(x)limx→a​f(x)​, provided that $\lim_{x\to a}g(x) \neq 0$limx→a​g(x)=0.

In other words, the limit of a quotient is the quotient of the limits, provided that the denominator is non-zero. So, we have:

$\lim_{x\to 3}\frac{2x+4}{x^3} = \frac{ \lim_{x\to 3}(2x+4)}{ \lim_{x\to 3}(x^3)} = \frac{2(3)+4}{3^3} = \frac{10}{27}$limx→3​x32x+4​=limx→3​(x3)limx→3​(2x+4)​=332(3)+4​=2710​

Exercise 5

Let $f(x) = \frac{x^2-4}{x^2-2x}$f(x)=x2−2xx2−4​. Evaluate $\lim_{x\to 2}f(x)$limx→2​f(x).

Simply plugging in $x= 2$x=2 to evaluate this limit gives us $\frac{4-4}{4-4} = \frac{0}{0}$4−44−4​=00​. Since $\frac{0}{0}$00​ is an indeterminate form, we’ll have to use a different technique.

Let’s try factoring the numerator and denominator and then cancel like terms.

$\lim_{x\to 2}\frac{x^2-4}{x^2-2x} = \lim_{x\to 2}\frac{(x-2)(x+2)}{x(x-2)}$limx→2​x2−2xx2−4​=limx→2​x(x−2)(x−2)(x+2)​

$= \lim_{x\to 2}\frac{x+2}{x}$=limx→2​xx+2​

$= \frac{2+2}{2}$=22+2​

$= \frac{4}{2}$=24​

$= 2$=2

Exercise 6

Let $f(x) = \frac{\sqrt{x}-4}{x-16}$f(x)=x−16x​−4​. Evaluate $\lim_{x\to 16}f(x)$limx→16​f(x).

Again, plugging $x = 16$x=16 into this limit gives us $\frac{0}{0}$00​. Since we can’t factor or simplify, we’ll try a new technique called rationalization. Rationalization involves multiplying the term with a radical by its conjugate. Multiplying by the conjugate is a clever way of multiplying by 1 and helps to simplify expressions with radicals. Using rationalization, we have:

$\lim_{x\to 16}\frac{\sqrt{x}-4}{x-16} = \lim_{x\to 16}\frac{\sqrt{x}-4}{x-16} \cdot \frac{\sqrt{x}+4}{\sqrt{x}+4}$limx→16​x−16x​−4​=limx→16​x−16x​−4​⋅x​+4x​+4​

$= \lim_{x\to 16}\frac{x-16}{(x-16)(\sqrt{x}+4)}$=limx→16​(x−16)(x​+4)x−16​

$= \lim_{x\to 16}\frac{1}{\sqrt{x}+4}$=limx→16​x​+41​

$= \frac{1}{4+4}$=4+41​

$= \frac{1}{8}$=81​

Exercise 7

Let $f(x) = x \sin{(\frac{1}{x})}$f(x)=xsin(x1​). Evaluate $\lim_{x\to 0 }f(x)$limx→0​f(x).

Factorization and rationalization can’t help us here. Instead, we'll use the Squeeze Theorem. The Squeeze Theorem states that if $f(x) \leq h(x) \leq g(x)$f(x)≤h(x)≤g(x) for all $x$x on $[a, b]$[a,b] (except possibly at $x = c$x=c), and $\lim_{x\to c}f(x) = \lim_{x\to c}g(x) = L$limx→c​f(x)=limx→c​g(x)=L, then $\lim_{x\to c}h(x) = L$limx→c​h(x)=L. In other words, if $g(x)$g(x) is sandwiched between $f(x)$f(x) and $h(x)$h(x)on $[a, b]$[a,b], and if $f(x)$f(x) and $h(x)$h(x) have the same limit $L$L at $c$c, then $g(x)$g(x) is caught between them and must also have the same limit $L$L at $c$c.

Basically, the Squeeze Theorem sandwiches a function that is hard to evaluate between two simpler functions.

Using the Squeeze Theorem and our knowledge of trigonometric functions, we have:

$-1 \leq \sin{\frac{1}{x}} \leq 1$−1≤sinx1​≤1

$-x\leq x\sin{(\frac1x)}\leq x$−x≤xsin(x1​)≤x

$\lim_{x\to 0 }-x = 0$limx→0​−x=0 and $\lim_{x\to 0 }x = 0$limx→0​x=0

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