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

\lim_{x\to\1^{-1}} \dfrac{x^2+2x-3}{|x-1|}

__Solution to Example 1:__

Note that we are looking for the limit as x approaches 1 from the left ( x → 1^{-1} means x approaches 1 by 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

\lim_{x\to\1^{-1}} \dfrac{(x-1)(x+3)}{-(x-1)}

Simplify to obtain

\lim_{x\to\1^{-1}} \dfrac{\cancel{(x-1)}(x+3)}{-\cancel{(x-1)}}

= \lim_{x\to\1^{-1}} -(x+3) = - 4

** Example 2:** Find the limit

\lim_{x\to\5} \dfrac{x^2-5}{x^2+x-30}

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.

= \lim_{x\to\5} \dfrac{(x-5)(x+5)}{(x-5)(x+6)}

Simplify to obtain

= \lim_{x\to\5} \dfrac{(x+5)}{(x+6)} = \dfrac{10}{11}

** Example 3:** Calculate the limit

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

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

** Example 4:** Calculate the limit

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

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.

** Example 6:** Find the limit

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.

** Example 7:** Find the limit

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

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

We first factor out 16 x

** Example 10:** Find the limit

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

Factor x

As x takes large values (infinity), the terms 2/x and 1/x

** Example 12:** Find the limit

Factor x

As x takes large values (infinity), the terms 1/x and 1/x

** Example 13:** Find the limit

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.

** Example 14:** Find the limit

Factor x

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

Multiply numerator and denominator by the conjugate and simplify

** Example 15:** Find the limit

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)

1) 3

2) 1

3) 1

4) 1/4

5) 0

6) 4

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