Informative line

Definition Of Limit

Learn definition of infinite limit calculus with examples, one sided limits, and vertical asymptotes. Practice to find limits from a graph.

Definition of Limit

  • Consider the function

\(f(x)=x^2+x+1\)

The values of \('f'\) for values of x which are near \(x = 1\), are as shown:

\(x\) \(f(x)\) \(x\) \(f(x)\)
.7 2.19 1.3 3.99
.8 2.44 1.2 3.64
.9 2.71 1.1 3.31
.95 2.8525 1.05 3.1525
.99 2.9701 1.01 3.031
.999 2.997 1.001 3.003

Observe that as \(x\) approaches close to 1, the values of \(f(x)\) approach 3. We write this as 

\(\lim\limits_{x\rightarrow 1}(x^2+x+1)=3\)    (lim is short for the Limit)

  • \(\lim\limits_{x\rightarrow a}f(x)=k\) will mean that as  \(x\) takes values closer and closer to 'a', \(f(x)\) will take values closer and closer to k.
  • We are not interested in what is happening at \(x=a\) in the above problem. In fact, function may not be defined at \(x=a\).

Illustration Questions

Find \(\lim\limits_{x\rightarrow 2}(x^2+2x+4)\) by taking values of x closer to 2 on both sides of x = 2.

A 12

B 8

C 1

D –4

×

Look at the table:

\(x\) \(f(x)=x^2+2x+4\) \(x\) \(f(x)=x^2+2x+4\)
1.7 10.29 2.3 13.89
1.8 10.84 2.2 13.24
1.9 11.41 2.1 12.61
1.99 11.94 2.01 12.06
1.999 11.994 2.001 12.006

Look at the values of \(f(x)\) at  x = 1.999 and x = 2.001, these values are very close to 12. So, we guess the value of limit is 12.

Find \(\lim\limits_{x\rightarrow 2}(x^2+2x+4)\) by taking values of x closer to 2 on both sides of x = 2.

A

12

.

B

8

C

1

D

–4

Option A is Correct

One Sided Limits

Note that  \(x\rightarrow2\)  means \(x\) takes values very close to 2, we now distinguish between the two cases when \(x\)  is close to '2' but higher  than '2' and lower than '2'.

So,

 \(x\rightarrow a^+\)   \(\Rightarrow\)  \(x\) takes values close to 'a' but higher than 'a'

 \(x\rightarrow a^-\)    \(\Rightarrow\) \(x\) takes values close to 'a' but lower than 'a'

So, 

\(x\rightarrow 3^-\) will mean \(x\) = 2.9999 and 

\(x\rightarrow 3^+\) will mean  \(x\) = 3.0001

  • \(\lim\limits_{x\rightarrow a^-} f(x)=L\) will mean that values of f(x) approach L. When we take values of  \(x\) closer to 'a' and lower than 'a' or simply the left hand limit of f(x) at \( x = a\)  is L.
  • Similarly, \(\lim\limits_{x\rightarrow a^+} f(x)=M\) will mean that right hand limit of f(x) at  \( x = a\)  is M.

Illustration Questions

Let  f(x) = –1   if   \(x\geq1 \)  ... (1) f(x) = 2     if   \(x<1 \)   ... (2)  The value of \(\lim\limits_{x\rightarrow 1^+} f(x)\) will be

A 5

B 7

C –1

D 8

×

\(x\rightarrow 1^+ \) means x = 1.001,

So, rule number (1) will apply.

\(\because\) \( f(x) = –1\)   if   \(x\geq1 \)   

\(\therefore\) \(\lim\limits_{x\rightarrow 1^+} f(x)=-1\)

image

Let  f(x) = –1   if   \(x\geq1 \)  ... (1) f(x) = 2     if   \(x<1 \)   ... (2)  The value of \(\lim\limits_{x\rightarrow 1^+} f(x)\) will be

A

5

.

B

7

C

–1

D

8

Option C is Correct

Reading the Values From the Graphs

Consider the graph of a function f(x):

\(\lim\limits_{x\rightarrow2^+}f(x)=3\) (because the height of graph to the immediate right of 2 is 3)

\(\lim\limits_{x\rightarrow2^-}f(x)=2\) (because the height of graph to the immediate left of 2 is 2).

\(f(2)=3\) (empty circle indicates that (2, 2) is not a part of the graph)

Illustration Questions

Consider the graph of a function \('f'\)  given below. The value of\( f(–1)\) is

A 2

B 1

C 5

D –7

×

Empty circle indicates that (–1, 1) is not a part of the graph and \(f(-1)=2\).

Consider the graph of a function \('f'\)  given below. The value of\( f(–1)\) is

image
A

2

.

B

1

C

5

D

–7

Option A is Correct

Existence of Limit

We say that 

\(\lim\limits_{x\rightarrow a}f(x)\) exists if

 \(\lim\limits_{x\rightarrow a^{-}}f(x)=\lim\limits_{x\rightarrow a^{+}}f(x)\)

or left hand limit at ''x = a" = right hand limit at ''x = a''.

To find the existence of the limit \(\lim\limits_{x\rightarrow a}f(x)\) we find

  • R.H.L. = \(\lim\limits_{x\rightarrow a^+}f(x)\) (look at the height of graph to immediate right of a )
  • L.H.L. = \(\lim\limits_{x\rightarrow a^-}f(x)\) (look at the height of graph to immediate left of a).

Compare both values, if they are not equal we say that limit doesn't exist. If they are equal we say limit exists and equals R.H.L. or L.H.L.

Illustration Questions

The graph of a function \('f'\) is shown in figure. Which of the following limit exists?

A \(\lim\limits_{x\rightarrow 1}f(x)\)

B \(\lim\limits_{x\rightarrow {-3}}f(x)\)

C \(\lim\limits_{x\rightarrow {3}}f(x)\)

D \(\lim\limits_{x\rightarrow {4}}f(x)\)

×

Consider option (A)

\(\lim\limits_{x\rightarrow {1^+}}f(x)=2\) (height of the graph at 1+)

\(\lim\limits_{x\rightarrow {1^-}}f(x)=1\) (height of the graph at 1)

Now, \(\lim\limits_{x\rightarrow {1^+}}f(x)\neq \lim\limits_{x\rightarrow {1^-}}f(x)\)

so,  \(\lim\limits_{x\rightarrow {1^+}}f(x)\) does not exist.

image

Consider option (B)

\(\lim\limits_{x\rightarrow {-3^+}}f(x)=0\) (height of the graph at –3+)

\(\lim\limits_{x\rightarrow {-3^-}}f(x)=0\) (height of the graph at –3)

Now, \(\lim\limits_{x\rightarrow {-3^+}}f(x)= \lim\limits_{x\rightarrow {-3^-}}f(x)\)

So, \(\lim\limits_{x\rightarrow {-3}}f(x)\)  exists.

image

Consider option (C)

\(\lim\limits_{x\rightarrow {3^+}}f(x)=2\) (height of the graph at 3+)

\(\lim\limits_{x\rightarrow {3^-}}f(x)=3\) (height of the graph at 3)

Now, \(\lim\limits_{x\rightarrow {3^+}}f(x)\neq \lim\limits_{x\rightarrow {3^-}}f(x)\)

So,  \(\lim\limits_{x\rightarrow {3}}f(x)\)  does not exist.

image

Consider option (D)

\(\lim\limits_{x\rightarrow {4^+}}f(x)=3\) (height of the graph at 4+)

\(\lim\limits_{x\rightarrow {4^-}}f(x)=1\) (height of the graph at 4)

Now, \(\lim\limits_{x\rightarrow {4^+}}f(x)\neq \lim\limits_{x\rightarrow {4^-}}f(x)\)

So,  \(\lim\limits_{x\rightarrow {4}}f(x)\)  does not exist.

image

The graph of a function \('f'\) is shown in figure. Which of the following limit exists?

image
A

\(\lim\limits_{x\rightarrow 1}f(x)\)

.

B

\(\lim\limits_{x\rightarrow {-3}}f(x)\)

C

\(\lim\limits_{x\rightarrow {3}}f(x)\)

D

\(\lim\limits_{x\rightarrow {4}}f(x)\)

Option B is Correct

Infinite Limits

We sometimes say,

\(\lim\limits_{x\rightarrow a}\;f(x)=\infty\), this means that values of f(x) can be made as large as possible when values of x approach 'a', or values of f(x) increase without bounds when \(x\rightarrow a\).

  • Note that \(\infty\) is not a number, its just a symbol, used to express a concept.
  • Similarly, \(\lim\limits_{x\rightarrow a}\;f(x)=-\infty\) will mean that values of f(x) can be made as large negative as possible when values of x approach 'a'.

In the graph shown:

Illustration Questions

Find the value of following infinite limit. \(\lim\limits_{x\rightarrow 7^+}\;\dfrac {5x}{x-7}\)

A \(\infty\)

B \(-\infty\)

C 2

D –3

×

\(x\rightarrow 7^+\Rightarrow x\) takes values close to 7 but more than 7 

\(\Rightarrow x-7\rightarrow 0^+ \)or very small positive value.

Now, numerator  \(5x\rightarrow35\) as \(x\rightarrow 7^+\)

so, \(\lim\limits_{x\rightarrow 7^+}\;\dfrac {5x}{x-7}=\dfrac {35}{0^+}=+\infty\)

In infinite limits denominators will tend to 0, you have to observe whether it is \(0^+\) or \(0^-\).

Find the value of following infinite limit. \(\lim\limits_{x\rightarrow 7^+}\;\dfrac {5x}{x-7}\)

A

\(\infty\)

.

B

\(-\infty\)

C

2

D

–3

Option A is Correct

Vertical Asymptotes

The vertical line x = a is called the vertical asymptote of a function y = f(x) or a curve y = f(x) if at least one of following six is true.

(1) \(\lim\limits_{x\rightarrow a^+}\;f(x)=\infty\)

(2) \(\lim\limits_{x\rightarrow a^-}\;f(x)=\infty\)

(3) \(\lim\limits_{x\rightarrow a^+}\;f(x)=-\infty\)

(4) \(\lim\limits_{x\rightarrow a^-}\;f(x)=-\infty\)

(5) \(\lim\limits_{x\rightarrow a}\;f(x)=\infty\)

 

(6) \(\lim\limits_{x\rightarrow a}\;f(x)=-\infty\)

  • To find the vertical asymptote of a curve y = f(x), put the denominator of f(x) = 0 and solve for x. If the solution is x = a then it is the vertical asymptote.

Illustration Questions

The vertical asymptotes of the function  \(f(x)=\dfrac {x+3}{(x^2-4x+3)}\)   is /are  

A \(x = 3, \,x = 1\)

B \(x = 4, \,x = 7\)

C \(x = –10, \,x = 8\)

D \(x = –5,\, x = 3\)

×

For vertical asymptote

Denominator = \(x^2-4x+3=0\)

\((x-3)(x-1)=0\)

\(\Rightarrow x=3,\; x=1\)

The vertical asymptotes of the function  \(f(x)=\dfrac {x+3}{(x^2-4x+3)}\)   is /are  

A

\(x = 3, \,x = 1\)

.

B

\(x = 4, \,x = 7\)

C

\(x = –10, \,x = 8\)

D

\(x = –5,\, x = 3\)

Option A is Correct

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