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How to Find Limits in Calculus

12.22.2021 • 12 min read

Rachel McLean

Subject Matter Expert

This article is a guide on how to find limits in calculus. We’ll learn several different methods for evaluating limits, and practice with some examples.

In This Article

  1. What Is a Limit?

  2. How to Find Limits

  3. Practice Exercises and Solutions

What Is a Limit?

Before getting into how to find a limit, let’s first define a limit. Given a function ff, a limit is the word used to describe the value that f(x)f(x) tends towards as xx tends towards some value. The notation for the limit of a function ff as xx approaches some number cc looks like this:

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

You can read the above notation as, “The limit of f(x)f(x) as xx approaches cc equals LL.”

In other words, f(x)f(x) tends towards LL as xx tends towards cc.

How to Find Limits

When finding limits, there are several methods we can use. We will go over six possible methods:

  • Direct Substitution

  • Factorization

  • Rationalization

  • The Squeeze Theorem

  • Trigonometric Identities

  • L'Hôpital's Rule

  • The method we use to find the limit, however, will depend on the type of function.

Direct Substitution

The term direct substitution is exactly as it sounds; you directly substitute a given value into a limit. Now, when do we need this method? If your function ff is continuous, the value of ff at cc and the limit of f(x)f(x) as xx approaches cc are the same. In other words, limxcf(x)=f(c)\lim_{x\to c}f(x) = f(c).

This rule is always true for polynomials, since polynomials are always continuous. Then, to evaluate a continuous function, we can simply substitute into f(x)f(x) the value at which we want to find the limit.

Example 1

For example, evaluate limx0(x2+x+10)\lim_{x\to 0 } (x^2 + x + 10). Since f(x)f(x) is a polynomial, we know that f(x)f(x) is continuous. So, we can simply substitute x=0x = 0 into f(x)f(x)!

limx0(x2+x+10)=02+0+10=10 \lim_{x\to 0 }(x^2 + x + 10) = 0^2 + 0 + 10 = 10

It’s worth noting that the value of ff at cc and the limit of f(x)f(x) as xx approaches cc are not always the same. We can often evaluate the limit of f(x)f(x) as xx approaches cc even if f(c)f(c) is undefined.

Example 2

For another example, consider the rational function f(x)=1x2f(x) = \frac{1}{x-2}. We can’t have a denominator of zero, so f(2)=122=10f(2)=\frac{1}{2-2}=\frac{1}{0} is undefined.

However, we can still evaluate the one-sided limits of f(x)f(x) as xx approaches 2, even though f(2)f(2) is undefined.

The term "one-sided limits" refers to:

  • The left-sided limit, limxcf(x)\lim_{x\to c^- }f(x). This is the limit as xx approaches some value cc from the left.

  • The right-sided limit, limxc+f(x)\lim_{x\to c^+ }f(x). This is the limit as xx approaches some value cc from the right.

Take a look at the graph of f(x)=1x2f(x) = \frac{1}{x-2} below.

CHART 1 How to Find Limits

Look closely at the behavior of the function near x=2x = 2. The variable xx moves infinitely closer to 2 from both sides, yet never reaches 2.

As xx approaches 2 from the left, yy gets infinitely smaller.

As xx approaches 2 from the right, yy gets infinitely bigger. Then, the following one-sided limits are true:

limx2+1x2= \lim_{x\to 2^+ }\frac{1}{x-2} = -\infty
limx21x2=+\lim_{x\to 2^- }\frac{1}{x-2} = +\infty

Since the one-sided limits approach ++ \infty and - \infty, we say there is a vertical asymptote at x=2x=2. For this example, the limit of f(x)f(x) as xx approaches 2 does not exist, since limx2+f(x)limx2f(x)\lim_{x\to 2^+ }f(x) \neq \lim_{x\to 2^- }f(x).

For other examples, the limit will exist where there is a vertical asymptote, provided that limxa+f(x)=limxaf(x)\lim_{x\to a^+ }f(x) = \lim_{x\to a^-}f(x). In order for a limit to exist, the one-sided limits must always equal each other. If the limit exists at a vertical asymptote, the value of that limit will always be ++ \infty or - \infty. Usually, if you substitute x=ax=a into f(x)f(x) and get that the denominator is zero while the numerator is non-zero, that’s a good sign that there’s a vertical asymptote at x=ax = a.

To give a more precise definition, we have this rule:

If any of the following statements are true, then x=ax=ais a vertical asymptote of f(x)f(x).

  • limxaf(x)=\lim_{​x\rightarrow a}f(x)=\infty

  • limxaf(x)=\lim_{​x\rightarrow a}f(x)=-\infty

  • limxa+f(x)=\lim_{​x\rightarrow a^+}f(x)=\infty

  • limxa+f(x)=\lim_{​x\rightarrow a^+}f(x)=-\infty

  • limxaf(x)=\lim_{​x\rightarrow a^-}f(x)=\infty

  • limxaf(x)=\lim_{​x\rightarrow a^-}f(x)=-\infty

Example 3

Let’s try direct substitution with one more example.

Let f(x)=x2+x6x2f(x) = \frac{x^2+x-6}{x-2}.

Evaluate limx2f(x)\lim_{x\to 2 }f(x).

Trying direct substitution, we find that f(2)=22+2622=660=00f(2) = \frac{2^2+2-6}{2-2} = \frac{6-6}{0} = \frac{0}{0}.

When evaluating limits, we say that 00\frac{0}{0} is indeterminate. This means that 00\frac{0}{0} does not give us a precise enough idea of how f(x)f(x) behaves as xx approaches 2.

Since we define division as simply the inverse of multiplication, we can say that n=00n = \frac{0}{0} implies that 0=n00 = n \cdot 0 for any real number nn.

Since any number multiplied by 0 is simply 0, there are an infinite number of answers. So, 00\frac{0}{0} is indeterminate and indicates that we’ll need to use a different method listed below.


When direct substitution gives us an indeterminate form like 00\frac{0}{0}, we can try factoring. Factoring allows us to cancel factors that are common to both the numerator and the denominator and then apply the Quotient Rule.

The Quotient Rule 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)}, given that limxag(x)0\lim_{x\to a}g(x) \neq 0. Said more simply, the limit of a quotient is the quotient of the limits, provided that the denominator is non-zero.

The Quotient Rule often allows us to apply direct substitution after factoring.

Let’s try factoring our last example, where we wanted to evaluate limx2x2+x6x2\lim_{x\to 2 }\frac{x^2+x-6}{x-2}.

limx2x2+x6x2=limx2(x+3)(x2)(x2)=limx2(x+3)1=(2+3)=5\lim_{x\to 2 }\frac{x^2+x-6}{x-2} = \lim_{x\to 2 }\frac{(x+3)(x-2)}{(x-2)} = \lim_{x\to 2 }\frac{(x+3)}{1} = (2+3) = 5

Thus, limx2x2+x6x2=5\lim_{x\to 2 }\frac{x^2+x-6}{x-2} = 5.


When dealing with limits that have radicals, it’s useful to move the radical from the numerator to the denominator or vice versa. To do this, we use rationalization. This technique involves multiplying by the conjugate. We can find the conjugate of an expression by switching the sign in the middle of the expression.

For example, let’s evaluate limx0x+11x\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x}. Since substituting in x=0x = 0 gives us 00\frac{0}{0}, we’ll try multiplying by the conjugate of the numerator.

limx0x+11x=limx0x+11xx+1+1x+1+1\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}
=limx0x+11x(x+1+1)= \lim_{x\to 0}\frac{x+1-1}{x(\sqrt{x+1}+1)}
=limx0xx(x+1+1)= \lim_{x\to 0}\frac{x}{x(\sqrt{x+1}+1)}
=limx01x+1+1) = \lim_{x\to 0}\frac{1}{\sqrt{x+1}+1)}
=10+1+1) = \frac{1}{\sqrt{0+1}+1)}
=12 = \frac{1}{2}

Thus, limx0x+11x=12\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac{1}{2}.

The Squeeze Theorem

The Squeeze Theorem is a method where we “sandwich” a function between two easier ones in order to evaluate its limit. This method states that if we can find two functions f(x)f(x) and h(x)h(x) that “squeeze” or trap g(x)g(x) between them on some interval [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) must also have the same limit LL at cc.

More precisely, 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

Again, the Squeeze Theorem traps one tricky function between two easier functions.

Using the Squeeze Theorem, let’s evaluate h(x)=limx0sin(x)xh(x) = \lim_{x\to 0}\frac{\sin (x)}{x}.

We need to trap h(x)h(x) between two easier functions. To find these functions, look at the graph below, which graphs f(x)=cos(x)f(x) = \cos (x), g(x)=1g(x) = 1, and h(x)=sin(x)xh(x) = \frac{\sin (x)}{x}.

How to Find Limits - graph with f(x)

Using this graph, we can say that cos(x)sin(x)x1\cos (x) \leq \frac{\sin (x)}{x} \leq 1.

Now, note that:

limx0cos(x)=1\lim_{x\to 0}\cos (x) = 1
limx01=1\lim_{x\to 0}1 = 1

Since h(x)=sin(x)xh(x) = \frac{\sin (x)}{x} is sandwiched between f(x)=cos(x)f(x) = \cos (x) and g(x)=1g(x) = 1, and since limx0f(x)=limx0g(x)=1\lim_{x\to 0}f(x) = \lim_{x\to 0}g(x) = 1, the Squeeze Theorem says that limx0sin(x)x=1\lim_{x\to 0}\frac{\sin (x)}{x} = 1.

Trigonometric Identities

Using trigonometric identities is another clever way to manipulate functions so that we can more easily evaluate the limit. Trigonometric identities are rules involving trigonometric functions that are always true. These equations can be substituted into trigonometric functions to make the problem easier.

While there are many different trigonometric identities, familiarizing yourself with them all is a worthwhile endeavor.

Some identities are for evaluating trigonometric functions are:

  • Pythagorean Identities

  • Reciprocal Identities

  • Double Angle Identities

  • Quotient Identities

For this example, we’ll look at a Pythagorean identity.

Let f(x)=1cos(x)xf(x) = \frac{1-\cos{(x)}}{x}. Let’s evaluate limx0f(x)\lim_{x\to 0 }f(x).

Plugging in x=0x=0 gives us f(x)=00f(x) = \frac{0}{0}, so we’ll have to find another way.

First, let’s multiply by the conjugate of the numerator. We can find the conjugate of the numerator by simply switching the sign in the middle of the expression. Multiplying by the conjugate is a very handy tool that allows us to discover trigonometric identities that aren’t obvious in the original form of the given function. In this problem, we’ll use a Pythagorean identity, which states that sin2(x)+cos2(x)=1\sin ^2 (x) + \cos ^2 (x) = 1. So, we have:

limx01cos(x)x=limx01cos(x)x1+cos(x)1+cos(x)\lim_{x\to 0}\frac{1-\cos{(x)}}{x} = \lim_{x\to 0}\frac{1-\cos{(x)}}{x} \cdot \frac{1+\cos{(x)}}{1+\cos{(x)}}
=limx01cos2(x)x(1+cos(x) = \lim_{x\to 0}\frac{1- \cos ^2 (x)}{x(1+ \cos (x)}
=​​limx0sin2(x)x(1+cos(x)= ​​\lim_{x\to 0}\frac{\sin ^2 (x)}{x(1+ \cos (x)}
=limx0sin(x)xsin(x)1+cos(x)= \lim_{x\to 0}\frac{\sin (x)}{x} \cdot \frac{\sin (x)}{1 + \cos (x)}

By the Product Rule, we know that:

limx0sin(x)xsin(x)1+cos(x)=limx0sin(x)xlimx0sin(x)1+cos(x)\lim_{x\to 0}\frac{\sin (x)}{x} \cdot \frac{\sin (x)}{1 + \cos (x)} = \lim_{x\to 0}\frac{\sin (x)}{x} \cdot \lim_{x\to 0}\frac{\sin (x)}{1 + \cos (x)}

And, using our answer from the Squeeze Theorem example, we know that limx0sin(x)x=1\lim_{x\to 0}\frac{\sin (x)}{x} = 1. Then, we have:

limx01cos(x)x=limx0sin(x)xlimx0sin(x)1+cos(x)\lim_{x\to 0}\frac{1-\cos{(x)}}{x} = \lim_{x\to 0}\frac{\sin (x)}{x} \cdot \lim_{x\to 0}\frac{\sin (x)}{1 + \cos (x)}
=1limx0sin(x)1+cos(x)= 1 \cdot \lim_{x\to 0}\frac{\sin (x)}{1 + \cos (x)}
=1sin(0)1+cos(0)= 1 \cdot \frac{\sin (0)}{1+ \cos (0)}
=102= 1 \cdot \frac{0}{2}
=0 = 0

Thus, limx01cos(x)x=0\lim_{x\to 0}\frac{1-\cos{(x)}}{x} = 0.

L'Hôpital's Rule

L'Hôpital's Rule states that limxf(x)g(x)=limxf(x)g(x)\lim_{x\to\infty}\frac{f(x)}{g(x)} = \lim_{x\to\infty}\frac{f’(x)}{g’(x)} if the first limit limxaf(x)g(x)\lim_{x\to a}\frac{f(x)}{g(x)} is indeterminate. This means that when direct substitution gives us an indeterminate form, we can differentiate the numerator and denominator, and then take the same limit again. 

For example, evaluate limx2x2x24\lim_{x\to 2}\frac{x-2}{x^2-4}. Direct substitution gives us 00\frac{0}{0}, so we’ll try L'Hôpital's Rule. Let f(x)f(x) be the numerator and g(x)g(x) be the denominator.

limx2x2x24=limx2f(x)g(x)\lim_{x\to 2}\frac{x-2}{x^2-4} = \lim_{x\to 2}\frac{f’(x)}{g’(x)}
=limx212x = \lim_{x\to 2}\frac{1}{2x}
=12(2) = \frac{1}{2(2)}
=14 = \frac{1}{4}

Thus, limx2x2x24=14\lim_{x\to 2}\frac{x-2}{x^2-4}=\frac{1}{4}.

Practice Exercises and Solutions

Now that we’ve seen how to evaluate limits, here are exercises where you can practice how to find the limit of a function.

Exercise 1

Evaluate limx11(x1)2\lim_{x\to 1}\frac{1}{(x-1)^2}.

What does this limit suggest?


When we plug x=1x = 1 into f(x)=1(x1)2f(x) = \frac{1}{(x-1)^2}, we find that f(1)=10f(1) = \frac{1}{0}. This is a good indication that there may be a vertical asymptote at x=1x = 1.

CHART 3 Find Limits in Calculus

Looking at the graph of f(x)f(x), we see that yy approaches ++\infty as xx approaches 1 from the left.

And, yy approaches ++\infty as xx approaches 1 from the right. Thus, limx11(x1)2=\lim_{x\to 1}\frac{1}{(x-1)^2} = \infty, and there is a vertical asymptote at x=1x = 1.

Exercise 2

Evaluate limx1x25x6x+1\lim_{x\to -1}\frac{x^2-5x-6}{x+1}.


Direct substitution gives us limx1x25x6x+1=00\lim_{x\to -1}\frac{x^2-5x-6}{x+1} = \frac{0}{0}, so we’ll try factoring. Then, we have:

limx1x25x6x+1=limx1(x6)(x+1)x+1\lim_{x\to -1}\frac{x^2-5x-6}{x+1} = \lim_{x\to -1}\frac{(x-6)(x+1)}{x+1}
=limx1(x6) = \lim_{x\to -1}(x-6)
=16 = -1 -6
=7 = -7

Exercise 3

Evaluate limx4x2x4\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}.


Direct substitution gives us 00\frac{0}{0}, so we’ll use rationalization to solve this problem. To rationalize the radical in the numerator, we’ll multiply by its conjugate.

limx4x2x4=limx4x2x4x+2x+2\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4} = \lim_{x\to 4}\frac{\sqrt{x}-2}{x-4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}
=limx4x4(x4)(x+2= \lim_{x\to 4}\frac{x-4}{(x-4)(\sqrt{x}+2}
=limx41x+2= \lim_{x\to 4}\frac{1}{\sqrt{x}+2}
=14 = \frac{1}{4}

Exercise 4

Evaluate limx0x2cos(1x)\lim_{x\to 0} x^2 \cos (\frac{1}{x}).


For this problem, we’ll use the Squeeze Theorem. We need to trap h(x)=x2cos(1x)h(x) = x^2 \cos (\frac{1}{x}) between two easier functions.

1cos(1x)1-1 \leq \cos (\frac{1}{x}) \leq 1
x2x2cos(1x)x2-x^2 \leq x^2\cos (\frac{1}{x}) \leq x^2

limx0(x2)=0\lim_{x\to 0}(-x^2) = 0 and limx0(x2)=0\lim_{x\to 0}(x^2) = 0

Since h(x)h(x) is sandwiched between f(x)=x2f(x) = -x^2 and g(x)=x2g(x) = x^2 and since limx0f(x)=limx0g(x)=0\lim_{x\to 0}f(x) =\lim_{x\to 0}g(x) = 0, then limx0x2cos(1x)=0\lim_{x\to 0} x^2 \cos (\frac{1}{x}) = 0.

Exercise 5

Evaluate limx0sin(x)sin(2x)\lim_{x\to 0} \frac{\sin (x)}{\sin (2x)}.


Direct substitution gives us 00\frac{0}{0}, so we’ll use trig identities to solve this problem. Specifically, it will be helpful to use a Double Angle identity, which states that sin(2x)=2sin(x)cos(x)\sin (2x) = 2 \sin (x) \cos (x).

limx0sin(x)sin(2x)=limx0sin(x)2sin(x)cos(x)\lim_{x\to 0} \frac{\sin (x)}{\sin (2x)} = \lim_{x\to 0} \frac{\sin (x)}{2\sin (x) \cos(x)}
=limx012cos(x)= \lim_{x\to 0} \frac{1}{2 \cos (x)}
=12(1)= \frac{1}{2(1)}
=12= \frac{1}{2}

Exercise 6

Evaluate limx3x327x3\lim_{x\to 3} \frac{x^3-27}{x-3}.


Direct substitution gives us 00\frac{0}{0}, so we’ll use L'Hôpital's Rule. Let f(x)f(x) be the numerator and g(x)g(x) be the denominator.

limx3x327x3=limx3f(x)g(x)\lim_{x\to 3}\frac{x^3-27}{x-3} = \lim_{x\to 3}\frac{f’(x)}{g’(x)}
=limx33x21 = \lim_{x\to 3}\frac{3x^2}{1}
=3(3)2= 3(3)^2
=27= 27

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