Find the limits of various functions using different methods. Several Examples with detailed solutions are presented. More exercises with answers are at the end of this page.
Example 1: Find the limit
Solution to Example 1:

Note that we are looking for the limit as x approaches 1 from the left (values smaller than 1). Hence
x < 1
x  1 < 0

If x  1 < 0 then
 x  1  =  (x  1)

Substitute  x  1  by  (x  1), factor the numerator to write the limit as follows

Simplify to obtain
=  4
Example 2: Find the limit
Solution to Example 2:

Although the limit in question is the ratio of two polynomials, x = 5 makes both the numerator and denominator equal to zero. We need to factor both numerator and denominator as shown below.

Simplify to obtain
= 10 / 11
Example 3: Calculate the limit
Solution to Example 3:

We need to look at the limit from the left of 2 and the limit from the right of 2. As x approaches 2 from the left
x  2 < 0 hence
x  2 = (x  2)

Substitute to obtain the limit from the left of 2 as follows
=  8

As x approaches 2 from the right x  2 > 0 hence
x  2 = x  2

Substitute to obtain the limit from the right of 2 as follows
= 8

The limit from the right of 2 and the limit from the left of 2 are not equal therefore the given limit DOES NOT EXIST.
Example 4: Calculate the limit
Solution to Example 4:

As x approaches 1, cube root x + 1 approaches 0 and ln (x+1) approaches  infinity hence an indeterminate form 0 . infinity

Let us rewrite the limit so that it is of the infinity/infinity indeterminate form.

We now use L'hopital's Rule and find the limit.
Example 5: Find the limit
Solution to Example 5:

As x gets larger x + 1 gets larger and e^(1/(x+1)1) approaches 0 hence an indeterminate form infinity.0

Let us rewrite the limit so that it is of the 0/0 indeterminate form.

Apply the l'hopital's theorem to find the limit.
=  1
Example 6: Find the limit
Solution to Example 6:

As x approaches 9, both numerator and denominator approach 0. Multiply both numerator and denominator by the conjugate of the numerator.

Expand and simplify.

and now find the limit.
= 1 / 6
Example 7: Find the limit
Solution to Example 7:

The range of the cosine function is.
1 <= cos x <= 1

Divide all terms of the above inequality by x, for x positive.
1 / x <= cos x / x <= 1 / x

Now as x takes larger values without bound (+infinity) both 1 / x and 1 / x approaches 0. Hence by the squeezing theorem the above limit is given by
Example 8: Find the limit
Solution to Example 8:

As t approaches 0, both the numerator and denominator approach 0 and we have the 0 / 0 indeterminate form. Hence the l'hopital theorem is used to calculate the above limit as follows
Example 9: Find the limit
Solution to Example 9:

We first factor out 16 x^{ 2} under the square root of the denominator and take out of the square root and rewrite the limit as

Since x approaches larger positive values (infinity)  x  = x. Simplify and find the limt.
= 3 / 4
Example 10: Find the limit
Solution to Example 10:

As x approaches 2 from the left then x  2 approaches 0 from the left or x  2 < 0. The numerator approaches 5 and the denominator approaches 0 from the left hence the limit is given by
Example 11: Find the limit
Solution to Example 11:

Factor x^{ 2} in the denominator and simplify.

As x takes large values (infinity), the terms 2/x and 1/x^{ 2} approaches 0 hence the limit is
= 3 / 4
Example 12: Find the limit
Solution to Example 12:

Factor x^{ 2} in the numerator and denominator and simplify.

As x takes large values (infinity), the terms 1/x and 1/x^{ 2} and 3/x^{ 2} approaches 0 hence the limit is
= 0 / 2 = 0
Example 13: Find the limit
Solution to Example 13:

Multiply numerator and denominator by 3t.

Use limit properties and theorems to rewrite the above limit as the product of two limits and a constant.

We now calculate the first limit by letting T = 3t and noting that when t approaches 0 so does T. We also use the fact that sin T / T approaches 1 when T approaches 0. Hence

The second limit is easily calculated as follows

The final value of the given limit is
Example 14: Find the limit
Solution to Example 14:

Factor x^{ 2} inside the square root and use the fact that sqrt(x^{2}) =  x .

Since x takes large values (infinity) then  x  = x. Hence the indeterminate form

Multiply numerator and denominator by the conjugate and simplify

Factor x out of the numerator and denominator and simplify

As x gets larger, the terms 1/x and 1/x^{2} approach zero and the limit is
= 1 / 2
Example 15: Find the limit
Solution to Example 15:

Let z = 1 / x so that as x get large x approaches 0. Substitute and calculate the limit as follows.
Exercises: Calculate the following limits
1.
2.
3.
4.
5.
6.
Solutions to Above Exercises:
1) 3
2) 1
3) 1
4) 1/4
5) 0
6) 4
More on limits
Calculus Tutorials and Problems