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 fff, a limit is the value that f(x)f(x)f(x) approaches as xxx approaches some value.

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

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Suppose we want to find the limit of fff at x=2x = 2x=2. We want to find the yyy-value that fff approaches as xxxgets infinitely close to 2. Looking at the graph, it’s clear that yyyapproaches 4 as xxx gets closer and closer to 2. In this case, we say that the limit as xxx approaches 2 is 4.

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

For example, consider the function f(x)=x21x1f(x) = \frac{x^2-1}{x-1}f(x)=x1x21, graphed below. Notice that there is a removable discontinuity called a hole at x=1x = 1x=1, since we can’t have a denominator of zero.

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Even though f(x)f(x)f(x) is not defined at 1, we can still use limits to understand the behavior of the function as xxx approaches 1. As xxx moves infinitely close to 111, we see that yyy 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.

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

So, we can say that the limit of fff as xxx approaches 1 is 2, even though the value of fff at 111 is undefined. Notice that the value of fff as xxx approaches 1 gets arbitrarily close to 2 from both directions. This means that if we approach 1 from either the negative or positive direction, yyy still approaches 2.

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

Limit Notation

If we want to evaluate the limit of a real-valued function fff as xxx approaches some real number ccc, we use the following notation:

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

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

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 xxx 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} } = LΔx0limΔxf(x+Δx)f(x)=L

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

Limits are also used to define integrals. The limit definition of the definite integral of fff on the interval [a,b][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} = Labf(x)dx=limni=1nf(c1)Δxi=L

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

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

Exercise 2

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

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Solution: The limit does not exist. In the positive direction, f(x)f(x)f(x) approaches 5 as xxx approaches 2. In the negative direction, f(x)f(x)f(x) approaches 7 as xxx approaches 2. So, we can say that limx2+f(x)=5\lim_{x\to 2^+}f(x) = 5limx2+f(x)=5 and limx2f(x)=7\lim_{x\to 2^-}f(x) = 7limx2f(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 -3f(x)=x2+4x3. Evaluate limx2f(x)\lim_{x\to 2}f(x)limx2f(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)limxacf(x)=climxaf(x)

The Constant Multiple Rule states that we can bring the constant ccc 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)limxaf(x)±g(x)=limxaf(x)±limxag(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 = alimxax=a

The limit of xxx as xxx approaches aaa is equal to aaa.

  • limxac=c\lim_{x\to a}c = climxac=c

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

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}3limx2x2+4x3=limx2x2+4limx2xlimx23

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

=9= 9=9

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

Exercise 4

Let f(x)=2x+4x3f(x) = \frac{2x+4}{x^3}f(x)=x32x+4. Evaluate limx3f(x)\lim_{x\to3}f(x)limx3f(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)}limxag(x)f(x)=limxag(x)limxaf(x), provided that limxag(x)0\lim_{x\to a}g(x) \neq 0limxag(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:

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}limx3x32x+4=limx3(x3)limx3(2x+4)=332(3)+4=2710

Exercise 5

Let f(x)=x24x22xf(x) = \frac{x^2-4}{x^2-2x}f(x)=x22xx24. Evaluate limx2f(x)\lim_{x\to 2}f(x)limx2f(x).

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

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)} limx2x22xx24=limx2x(x2)(x2)(x+2)

=limx2x+2x = \lim_{x\to 2}\frac{x+2}{x} =limx2xx+2

=2+22 = \frac{2+2}{2} =22+2

=42 = \frac{4}{2} =24

=2 = 2=2

Exercise 6

Let f(x)=x4x16f(x) = \frac{\sqrt{x}-4}{x-16}f(x)=x16x4. Evaluate limx16f(x)\lim_{x\to 16}f(x)limx16f(x).

Again, plugging x=16x = 16x=16 into this limit gives us 00\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:

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} limx16x16x4=limx16x16x4x+4x+4

=limx16x16(x16)(x+4) = \lim_{x\to 16}\frac{x-16}{(x-16)(\sqrt{x}+4)} =limx16(x16)(x+4)x16

=limx161x+4 = \lim_{x\to 16}\frac{1}{\sqrt{x}+4} =limx16x+41

=14+4= \frac{1}{4+4}=4+41

=18 = \frac{1}{8}=81

Exercise 7

Let f(x)=xsin(1x)f(x) = x \sin{(\frac{1}{x})}f(x)=xsin(x1). Evaluate limx0f(x)\lim_{x\to 0 }f(x)limx0f(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)f(x)h(x)g(x) for all xxx on [a,b][a, b][a,b] (except possibly at x=cx = cx=c), and limxcf(x)=limxcg(x)=L\lim_{x\to c}f(x) = \lim_{x\to c}g(x) = Llimxcf(x)=limxcg(x)=L, then limxch(x)=L\lim_{x\to c}h(x) = Llimxch(x)=L. In other words, if g(x)g(x)g(x) is sandwiched between f(x)f(x)f(x) and h(x)h(x)h(x)on [a,b][a, b][a,b], and if f(x)f(x)f(x) and h(x)h(x)h(x) have the same limit LLL at ccc, then g(x)g(x)g(x) is caught between them and must also have the same limit LLL at ccc.

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 11sinx11

xxsin(1x)x-x\leq x\sin{(\frac1x)}\leq xxxsin(x1)x

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

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