Solving Limits Made Simple: Solve Any Limit You May Find

Here you'll find everything you need to know about solving limits. I prepared a list of all possible cases of limits. If you master these, there won't be a single limit you can't solve.

My goal for this page is to be the ultimate resource for solving limits. It may be a good idea to bookmark it. You'll find solved examples and tips for every type of limit.

Remember that you can post your questions here: Your Limits Questions. If you aren't able to solve a problem even after going through this page, you may post a question about it there.

One thing is to understand intuitively what limits are, another is to know how to solve any limit problem you might face.

To get the intuition of what limits are (I really recommend to get that before learning all the techniques presented here), you can visit the pages:

• Limits and Continuity: Here you'll find an intuitive approach to the concepts of limits and continuity. This is a good starting point.
• Limits of Functions: Here you'll go deeper into the concept of limits of functions. You'll also learn the basic properties of limits, which are used in any problem that involves solving limits.

Type 1: Limits By Simple Substitution

These are the easiest types of limits, and usually the first thing you learn. In these limits you only need to substitute the value to which the independent variable is approaching. For example: Here you simply replace x by a to get: I don't think you need much practice solving these. They're not much fun either. However, there is an interesting question here by a reader that relates the technique we use here and the concept of continuity: Solving Limits by Continuity.

If you haven't learned about continuous functions yet, you can skip that page.

Type 2: Limits by Factoring

Now this is more interesting. In these limits, if you try to substitute as in the previous case, you get an indetermination. For example: If you simply substitute x by 1 in the expression you'll get 0/0. So, what can we do? We use our algebraic skills to simplify the expression. In the previous example we can factor the numerator: And that's it. There are not many secrets in these limits. It is also easy to spot them: whenever you see a quotient of two polynomials, you may try this technique if there is an indetermination.

Solving limits by factorization just requires you to remember your algebra days. Watch this video for more examples:

Type 3: Limits by Rationalization

These involve limits with square roots. Watch the video for examples:


In these limits we apply an algebraic technique called rationalization. For example: If we substitute we get 0/0 and we cannot factor this. The trick is to multiply and divide the fraction by a convenient expression. (Remember that if you multiply and divide a number by the same thing you get the same number). In this case we use the following identity: (Just perform the product in the left to verify it). So, whenever you see the difference or the sum of two square roots, you can apply the previous identity. The two factors in the left are called conjugate expressions.

In the example above, the conjugate of the numerator is: And that's the number we'll be multiplying and dividing our fraction by: Now, in the numerator we use the algebraic identity I just mentioned: Now the (1-x) goes away and we get the desired result: In these limits you always do the same. You recognize the difference between two square roots and the multiply and divide by the conjugate of the expression. Another example is: You get an indetermination if you substitute h by zero. You can see there the difference between two square roots in the numerator.

All you need to do is to multiply and divide by the conjugate of the numerator and work algebraically.

Here's another worked out example: Limit by Rationalization. There are other examples that are trickier, in the sense that you need to multiply by two expressions. For example: (p and q are constants). In this case you have square roots both on the numerator and denominator.

In this case you need to multiply and divide by two factors: the conjugate of the numerator and then the conjugate of the denominator.

This problem is good practice and I recommend you to try it. If you tried and still can't solve it, you can post a question about it together with your work.

Type 4: Limits at Infinity

In these limits the independent variable is approaching infinity. An example is the limit: I've already written a very popular page about this technique, with many examples: Solving Limits at Infinity. At the following page you can find also an example of a limit at infinity with radicals. In this limit you also need to apply the techniques of rationalization we've seen before: Limit with Radicals

Type 5: Trigonometric Limits

In most limits that involve trigonometric functions you must apply the fundamental limit: Watch the video for examples:


With things involving trigonometric functions you always need practice, because there are so many trigonometric identities to choose from.

In the following page you'll find everything you need to know about trigonometric limits, including many examples: The Squeeze Theorem and Limits With Trigonometric Functions.

Here also more examples of trigonometric limits. I think you'll find all techniques you need to know in these:

• Trigonometric Limit with tangent
• A Somewhat Different Trigonometric Limit

Type 6: Limits Involving Number e

Number e is defined as the following limit: There are some limits that can be solved using this fundamental limit. This is similar to what we do with trigonometric limits. We try to accomodate the function algebraically to apply the limit we already know. For example, the limit: At this page you can find a more elaborate example: Limit at Infinity Involving Number e.

Limits by L'Hopital's Rule

If you haven't learned about derivatives yet, you can skip this section. With L'Hopital's Rule we can solve limits using our skills for finding derivatives.

This rule says that to find the limit of a quotient, you only need to find the derivatives of both the numerator and denominator and apply the limit again.

This works only if the quotient is an indeterminate form 0/0 or infinity over infinity. For example: Both the numerator and denominator approach infinity. So, this is an indetermination of the form infinity over infinity. We derive the numerator and denominator and apply the limit again. That is: This is cool, isn't it? I'll be posting a page dedicated to this rule shortly. However, there is already a very interesting example posted by a reader that is very challenging here: Hairy Limit With L'Hopital's Rule.

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