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.

__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\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}

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

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.

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.

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.

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

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

We first factor out 16 x

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

Factor x

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

Factor x

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

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.

Factor x

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

Multiply numerator and denominator by the conjugate and simplify

Let z = 1 / x so that as x get large x approaches 0. Substitute and calculate the limit as follows.

1)

2)

3)

4)

5)

6)

1) 3

2) 1

3) 1

4) 1/4

5) 0

6) 4

Limits of Absolute Value Functions Questions

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