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Calculus

What Are Limits in Calculus?

12.09.2021 • 8 min read

Rachel McLean

Subject Matter Expert

This article is a brief guide on limits in calculus. We’ll discuss the definition of limits, explain how limits are used in calculus, and learn several different techniques for evaluating limits.

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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 ff, a limit is the value that f(x)f(x) approaches as xx approaches some value.

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

CHART 1 Limits in Calculus

Suppose we want to find the limit of ff at x=2x = 2. We want to find the yy-value that ff approaches as xxgets infinitely close to 2. Looking at the graph, it’s clear that yyapproaches 4 as xx gets closer and closer to 2. In this case, we say that the limit as xx approaches 2 is 4.

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

For example, consider the function f(x)=x21x1f(x) = \frac{x^2-1}{x-1}, graphed below. Notice that there is a removable discontinuity called a hole at x=1x = 1, since we can’t have a denominator of zero.

CHART 2 Limits in Calculus

Even though f(x)f(x) is not defined at 1, we can still use limits to understand the behavior of the function as xx approaches 1. As xx moves infinitely close to 11, we see that yy 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.

x
f(x)
0.9
f(0.9) = 1.9
0.99
f(0.99) = 1.99
0.999
f(0.999) = 1.999
0.9999
f(0.9999) = 1.9999
1
f(1) = undefined
1.0001
f(1.0001) = 2.0001
1.001
f(1.001) = 2.001
1.01
f(1.01) = 2.01
1.1
f(1.1) = 2.1

So, we can say that the limit of ff as xx approaches 1 is 2, even though the value of ff at 11 is undefined. Notice that the value of ff as xx approaches 1 gets arbitrarily close to 2 from both directions. This means that if we approach 1 from either the negative or positive direction, yy still approaches 2.

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

Limit Notation

If we want to evaluate the limit of a real-valued function ff as xx approaches some real number cc, we use the following notation:

limxcf(x)=L \lim_{x\to c}f(x) = L

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

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 xx if the following limit exists:

limΔx0f(x+Δx)f(x)Δx=L\mathop {\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x \right)}}{\Delta{x} } = L

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

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

abf(x)dx=limni=1nf(c1)Δxi=L\int_{a}^{b} f(x) \,dx = \lim_{n\to\infty}\sum_{i=1}^{n}f(c_1)\Delta{x_i} = L

In this case, the limit LL represents the area under the curve f(x)f(x) on the interval between aa and bb. 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 ff below, what is a good estimate for limx1f(x)\lim_{x\to 1}f(x)?

CHART 3 Limits in Calculus

Solution: limx1f(x)=1\lim_{x\to 1}f(x) = 1. Although f(1)=3f(1) = 3, the graph indicates that f(x)f(x) approaches 1 on both sides as xx approaches 1. This is an example of a scenario when the limit of a function at xx does not equal the value of the function at xx.

Exercise 2

Using the graph of ffbelow, what is a good estimate for limx2f(x)\lim_{x\to 2}f(x)?

CHART 4 Limits in Calculus

Solution: The limit does not exist. In the positive direction, f(x)f(x) approaches 5 as xx approaches 2. In the negative direction, f(x)f(x) approaches 7 as xx approaches 2. So, we can say that limx2+f(x)=5\lim_{x\to 2^+}f(x) = 5 and limx2f(x)=7\lim_{x\to 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)=x2+4x3f(x) = x^2 + 4x -3. Evaluate limx2f(x)\lim_{x\to 2}f(x).

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

  • limxacf(x)=climxaf(x)\lim_{x\to a}cf(x) = c\lim_{x\to a}f(x)

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

  • limxaf(x)±g(x)=limxaf(x)±limxag(x)\lim_{x\to a}f(x) \pm g(x) = \lim_{x\to a}f(x) \pm \lim_{x\to a}g(x)

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

  • limxax=a\lim_{x\to a}x = a

The limit of xx as xx approaches aa is equal to aa.

  • limxac=c\lim_{x\to a}c = c

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

Using these properties, we have:

limx2x2+4x3=limx2x2+4limx2xlimx23\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

=(2)2+4(2)3= (2)^2 + 4(2) - 3

=9= 9

Notice that in this example, f(2)=limx2f(x)f(2) = \lim_{x\to 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 limxaf(x)=f(a)\lim_{x\to a}f(x) = f(a), since polynomials are always continuous.

Exercise 4

Let f(x)=2x+4x3f(x) = \frac{2x+4}{x^3}. Evaluate limx3f(x)\lim_{x\to3}f(x).

For this problem, we can use the Quotient Rule, which states that limxaf(x)g(x)=limxaf(x)limxag(x)\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{ \lim_{x\to a}f(x)}{ \lim_{x\to a}g(x)}, provided that limxag(x)0\lim_{x\to a}g(x) \neq 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:

limx32x+4x3=limx3(2x+4)limx3(x3)=2(3)+433=1027 \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}

Exercise 5

Let f(x)=x24x22xf(x) = \frac{x^2-4}{x^2-2x}. Evaluate limx2f(x)\lim_{x\to 2}f(x).

Simply plugging in x=2x= 2 to evaluate this limit gives us 4444=00\frac{4-4}{4-4} = \frac{0}{0}. Since 00\frac{0}{0} 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.

limx2x24x22x=limx2(x2)(x+2)x(x2) \lim_{x\to 2}\frac{x^2-4}{x^2-2x} = \lim_{x\to 2}\frac{(x-2)(x+2)}{x(x-2)}
=limx2x+2x = \lim_{x\to 2}\frac{x+2}{x}
=2+22 = \frac{2+2}{2}
=42 = \frac{4}{2}
=2 = 2

Exercise 6

Let f(x)=x4x16f(x) = \frac{\sqrt{x}-4}{x-16}. Evaluate limx16f(x)\lim_{x\to 16}f(x).

Again, plugging x=16x = 16 into this limit gives us 00\frac{0}{0}. 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:

limx16x4x16=limx16x4x16x+4x+4 \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}
=limx16x16(x16)(x+4) = \lim_{x\to 16}\frac{x-16}{(x-16)(\sqrt{x}+4)}
=limx161x+4 = \lim_{x\to 16}\frac{1}{\sqrt{x}+4}
=14+4= \frac{1}{4+4}
=18 = \frac{1}{8}

Exercise 7

Let f(x)=xsin(1x)f(x) = x \sin{(\frac{1}{x})}. Evaluate limx0f(x)\lim_{x\to 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)h(x)g(x)f(x) \leq h(x) \leq g(x) for all xx on [a,b][a, b] (except possibly at x=cx = c), and limxcf(x)=limxcg(x)=L\lim_{x\to c}f(x) = \lim_{x\to c}g(x) = L, then limxch(x)=L\lim_{x\to 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 LL at cc, then g(x)g(x) is caught between them and must also have the same limit LL at cc.

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:

1sin1x1-1 \leq \sin{\frac{1}{x}} \leq 1

xxsinxxx-x \leq x\frac{\sin{x}}{x} \leq x

limx0x=0\lim_{x\to 0 }-x = 0 and limx0x=0\lim_{x\to 0 }x = 0

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