Explainer: Limits by Direct Substitution

In this explainer, we will learn how to use the direct substitution method to evaluate limits.

Recall that we are given a function 𝑓 defined near π‘Ž, and seek a number 𝐿 so that the closer π‘₯ gets to π‘Ž, the closer the values 𝑓(π‘₯) get to 𝐿. Such a number need not even exist. When it does, we write limο—β†’οŒΊπ‘“(π‘₯)=𝐿.

The substitution method says that 𝐿=𝑓(π‘Ž), which obviously requires that the function be defined at π‘Ž. So we cannot use this to find limο—β†’οŠ©23βˆ’π‘₯, just because the function 𝑓(π‘₯)=23βˆ’π‘₯ is not defined at π‘₯=3.

Even if 𝑓(π‘Ž) makes sense, substitution can still fail, depending on what the function is. Consider 𝐹(π‘₯)=π‘₯+1π‘₯β‰₯0,π‘₯βˆ’1π‘₯<0.ifif

Now 𝐹(0)=0+1=1, because π‘₯=0 satisfies the first condition: 0β‰₯0. In spite of this, we cannot say that limο—β†’οŠ¦πΉ(π‘₯)=0 because of the piecewise definition and the fact that this definition gives different one-sided limits to the left and right of the point π‘Ž=0, as clear from the graph.

We see that limο—β†’οŠ¦οŽͺ𝐹(π‘₯)=βˆ’1 while limο—β†’οŠ¦οŽ©πΉ(π‘₯)=1, meaning that the limit does not exist.

On the positive side, we list when we can apply substitution.

Conditions When Substitution Works

Given a function 𝑓 and a point π‘₯=π‘Ž on the number line, where 𝑓 is defined, we can set the limit as π‘₯ approaches π‘Ž to be just 𝑓(π‘Ž) under any of the following conditions:

  1. 𝑓 is a polynomial, which includes any constant function.
  2. 𝑓 is rational function which is defined at π‘₯=π‘Ž. In other words, 𝑓(π‘₯)=𝑃(π‘₯)𝑄(π‘₯), with 𝑃,𝑄 polynomials, and 𝑄(π‘Ž)β‰ 0.
  3. 𝑓 belongs to one of the β€œstandard” classes of functions: trigonometric, exponential, and logarithmic.
  4. 𝑓 is a power function, so 𝑓(π‘₯)=π‘₯ with 𝑝 any real number. This includes functions like 𝑓(π‘₯)=√π‘₯ since this is just π‘₯.
  5. 𝑓 is a constant multiple, sum, difference, product, or quotient of functions for which substitution works.
  6. 𝑓 is composed of functions for which substitution works, so 𝑓(π‘₯)=𝑔(β„Ž(π‘₯)), where β„Ž permits substitution at π‘₯=π‘Ž and 𝑔 allows substitution at π‘₯=β„Ž(π‘Ž).

Although this is quite a large number, we can use these results even when the function only has a piecewise definition, with the proper choice of the point π‘Ž.

The figure shows the same function 𝐹 above and π‘₯=1.5. Since 1.5>0, we have 𝐹(1.5)=1.5+1=2.5 and, in this case, it is still the case that limο—β†’οŠ§οŽ–οŠ«πΉ(π‘₯)=𝐹(1.5)=2.5 because the function 𝐹 agrees with another function 𝐺 near π‘₯=π‘Ž, where 𝐺 belongs to the list above.

  • Which function 𝐺? The polynomial function 𝐺(π‘₯)=π‘₯+1.
  • What does β€œnear” mean? This can be interpreted to mean on the interval (βˆ’1,2) which contains our point π‘Ž=1.5. On this interval, 𝐹=𝐺.
  • How does π‘₯=1.5 differ from π‘₯=0? The difference is that no matter what interval is chosen around 0, there is no single polynomial that defines 𝐹 on that interval.

Example 1: Finding the Limit of a Polynomial by Substitution

Determine limο—β†’οŠ±οŠ«οŠ¨ο€Ήβˆ’9π‘₯βˆ’6π‘₯βˆ’9.


The limit is the value of βˆ’9π‘₯βˆ’6π‘₯βˆ’9 at π‘₯=βˆ’5 because this is a polynomial function: limο—β†’οŠ±οŠ«οŠ¨οŠ¨ο€Ήβˆ’9π‘₯βˆ’6π‘₯βˆ’9=βˆ’9(βˆ’5)βˆ’6(βˆ’5)βˆ’9=βˆ’204.

Example 2: Finding the Limit of Root Functions at a Point by Direct Substitution

Determine limο—β†’οŠ―οŠ¨βˆš4π‘₯βˆ’9π‘₯+1.


The function 𝑓(π‘₯)=√4π‘₯βˆ’9π‘₯+1 is a composition of functions √4π‘₯βˆ’9π‘₯+1=𝑔(β„Ž(π‘₯)), where 𝑔(π‘₯)=√π‘₯=π‘₯,β„Ž(π‘₯)=4π‘₯βˆ’9π‘₯+1.(apowerfunction)(apolynomialfunction)

Since β„Ž(9)=244, 𝑓(9) is defined and equals √244=2√61. As a composition of functions that permit substitution, it follows that limο—β†’οŠ―οŠ¨βˆš4π‘₯βˆ’9π‘₯+1=𝑓(9)=2√61.

A more subtle example is the following.

Example 3: Finding the Limit of a Function Involving Absolute Values

Given 𝑓(π‘₯)=|π‘₯+11|βˆ’|π‘₯βˆ’18|, find lim→οŠͺ𝑓(π‘₯).


One way to think of the absolute value function is |π‘₯|=√π‘₯, meaning that it is a composition of functions suitable for using substitution to find limits.

The function 𝑓 is defined for any real number, including π‘₯=4, and is a difference of |π‘₯+11| and |π‘₯βˆ’18|, both of which are compositions of the absolute value function |β‹…| and polynomial functions.

So we can evaluate the limit using substitution: lim→οŠͺ𝑓(π‘₯)=𝑓(4)=|4+11|βˆ’|4βˆ’18|=15βˆ’14=1.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.