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Home » Calculus » Understanding the Concept of Limits

Understanding the Concept of Limits

We know division by zero is not possible in mathematics. If we consider the function definition as

f(x)=\dfrac{(x^{2}-1)}{(x-1)}

The value of f(x) at x=1 is indeterminate.

f(1)=\dfrac{(1^{2}-1)}{(1-1)} = \dfrac 0 0

More simply, the value of the function f(x) does not exist at x=1. So, instead of x=1 we consider values of x sufficiently close to 1, i.e., as close to 1 as possible.

x f(x)
0.5 1.50000
0.9 1.90000
0.99 1.99000
0.999 1.99900
0.9999 1.99990
0.99999 1.99999
… …

From the table above we can see that as the values of x approach 1, the value of the function f(x) approach 2. In the table we have stopped at 0.99999, but if we take values of x even closer to 1, the corresponding values of f(x) will be even closer to 2. Here, as the values of x increase towards 1 the values of f(x) increase towards 2. This is symbolically written as:

\lim\limits_{x \to 1+} f(x) = 2

It reads, “limit x tends to 1 plus f(x) is equal to 2”. It is to be noted that the ‘+’ sign signifies values of x greater than 1 and not “positive” values of x.

Let us now look at the following table

x f(x)
1.5 2.50000
1.1 2.10000
1.01 2.01000
1.001 2.00100
1.0001 2.00010
1.00001 2.00001
… …

Again from the above table we can see that as the values of x come closer and closer to 1, the corresponding values of f(x) come closer and closer to 2. In other words, as the values of x approach 1, the corresponding values of f(x) approach 2. The only difference is that here the values of x decrease towards 1 and the values of f(x) decrease towards 2. This is symbolically written as:

\lim\limits_{x \to 1-} f(x) = 2

It reads, “limit x tends to 1 minus f(x) is equal to 2”. It is to be noted that the ‘-‘ sign signifies values of x less than 1 and not “negative” values of x.

In practice, the numerical difference between the value x=1 and a value of x sufficiently close to 1 (such as, x=1.0000001 or x=0.9999999) can me made as small as we please and hence can be neglected. Similarly, the numerical difference between the value f(x)=2 and a value of f(x) very close to 2 can be made as small as we please and hence be neglected if the value of x is taken sufficiently close to 1.

In general, when the value of f(x) cannot be determined for a particular value of x, say, x=a then there may exist a definite finite number b, such that the value of f(x) gradually tends to that finite number b when x tends to a. However we cannot say whether that finite number b will always exist or not. From this observation, the concept of limit has been developed by mathematicians.

Limit of a Variable

Let us consider a real variable x. Let a be a constant. Then by ‘x tends to a’ we mean x successively assumes values either greater than or less than a and the numerical difference between the assumed value of x, i.e., |x-a| becomes smaller and smaller. In this case, x becomes very close to a (but x≠a) and we say ‘x approaches a’.

If x approaches a assuming values greater than a then we say ‘x tends to a from the right side’ and we denote it by x \to a^+.

If x approaches a assuming values less than a then we say ‘x tends to a from the left side’ and we denote it by x \to a^-.

Limiting Value of a Function

We assume x to be a real variable, a is a real constant and f(x) is a single-valued function of x.

If x gradually approaches a assuming values which are greater than a and if the corresponding values of f(x) exist and these values gradually approach a finite constant l_1, then $latexl_1$ is called the right hand limiting value of f(x) or the right hand limit of f(x) and it is denoted by,

\lim\limits_{x \to a+} f(x) = l_1

Again, if x gradually approaches a assuming values which are less than a and if the corresponding values of f(x) exist and these values gradually approach a finite constant $latexl_2$, then $latexl_2$ is called the left hand limiting value of f(x) or the left hand limit of f(x) and it is denoted by,

\lim\limits_{x \to a-} f(x) = l_2

When x approaches a assuming values either greater than or less than a and f(x) assumes finite values for every value of x and if those values of f(x) gradually approach a finite constant l, then l is called the limiting value of f(x). It is denoted by,

\lim\limits_{x \to a} f(x) = l

The limit of the function l = \lim\limits_{x \to a} f(x) exists only if both l_1 (the right hand limit) and l_2 (the left hand limit) exist and l_1=l_2, i.e., if

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

l does not exist if,

  • l_1 is indeterminate OR
  • l_2 is indeterminate OR
  • l_1 \neq l_2.

What do we mean by x \to \infty and x \to-\infty?

If a real variable x assumes positive values and increases without limit, taking up values larger than any large number one can imagine, then we say that the variable x tends to infinity in the positive direction and denote it by x \to \infty.

If a real variable x assumes negative values and increases numerically without limit, taking up values which are numerically larger than any large number one can imagine, then we say that the variable x tends to infinity in the negative direction and denote it by x \to-\infty.

Some Important Limits

1. If n is a rational number, then \lim\limits_{x \to a} \dfrac{x^n-a^n}{x-a} = n . a^{n-1}

2. If n is a rational number, then \lim\limits_{x \to 0} \dfrac{(1+x)^n-1}{x} = n

3. \lim\limits_{x \to 0} \dfrac{sin x}{x} = 1

4. \lim\limits_{x \to 0} \dfrac{e^x -1}{x} = 1

5. \lim\limits_{x \to 0} {(1+x)}^{\dfrac {1}{x}} = e

6. \lim\limits_{x \to 0} \dfrac{a^x -1}{x} = ln a [a>0]

7. \lim\limits_{x \to \infty} {(1+\dfrac{1}{x})}^x = e

Points to Remember

  • x \to 0^+ does not mean that x takes positive values. It means that x approaches 0 assuming values greater than 0.
  • x \to 0^- does not mean that x takes negative values. It means that x approaches 0 assuming values less than 0.
  • \lim\limits_{x \to a+} f(x) is called the right hand limit of f(x) at x=a and \lim\limits_{x \to a-} f(x) is called the left hand limit of f(x) at x=a.
  • \lim\limits_{x \to a} f(x) exists if \lim\limits_{x \to a+} f(x) and \lim\limits_{x \to a-} f(x) both exists and \lim\limits_{x \to a+} f(x) = \lim\limits_{x \to a-} f(x)
  • Limit of a function may not exist as and when the above conditions are not fulfilled. If not, the limit of the function at that point does not exist.

Questions and Answers

Question 1: Does \lim\limits_{x \to 2} \dfrac{x^2-4}{x-2} exist? If so find its value?

Solution: Let f(x) = \dfrac{x^2-4}{x-2}

x f(x)
1.9 3.9
1.99 3.99
1.999 3.999
1.9999 3.9999
… …
2.1 4.1
2.01 .401
2.001 4.001
… …

From the above table we can see that as x approaches the finite value 2 (assuming values either greater than or less than 2 and sufficiently close to 2) the values of f(x) gradually approach 4 and the difference between the values of f(x) and 4 can be made as small as we please. Hence, f(x) \to 4 as x \to 2, but x≠2 because f(x) is undefined at x=2.

  1. Left hand limit = \lim\limits_{x \to 2-} f(x) = 4
  2. Right hand limit= \lim\limits_{x \to 2+} f(x) = 4

Clearly both Right hand limit and left hand limit exist and are equal.

Therefore, \lim\limits_{x \to 2} \dfrac{x^2-4}{x-2} exists and \lim\limits_{x \to 2} \dfrac{x^2-4}{x-2} = 4

Question 2: Evaluate \lim\limits_{x \to 1} \dfrac{1}{(x-1)^2}

Solution: Let f(x) = \dfrac{1}{(x-1)^2}^2

x f(x)
0.9 10^2
0.99 10^4
0.999 10^6
… …
1.1 10^2
1.01 10^4
1.001 10^6
… …

From the above table we can see that as x gradually approaches the finite value 1 from the left, assuming values less than 1, the value of f(x) keeps increasing and approaches a large number, as large as we can imagine.

Left hand limit = \lim\limits_{x \to 1-} f(x) = +\infty

Also, we can see that as x gradually approaches the finite value 1 from the right, assuming values greater than 1, the value of f(x) keeps increasing and approaches a large number, as large as we can imagine.

Right hand limit = \lim\limits_{x \to 1+} f(x) = +\infty

So left hand limit is equal to right hand limit. \lim\limits_{x \to 1-} f(x) = \lim\limits_{x \to 1+} f(x) = +\infty

Therefore, \lim\limits_{x \to 1+} f(x) = \lim\limits_{x \to 1} \dfrac{1}{(x-1)^2} = +\infty

NOTE: Here, x≠1 because at x=1, f(x)=\dfrac {1}/{(1-1)^2} = \dfrac {1}{0} which is undefined.

Question 3: Evaluate \lim\limits_{x \to 0} \dfrac{1}{x}

Solution : Let f(x) = \dfrac{1}{x}

x f(x)
-0.1 -10
-0.01 -100
-0.001 -1000
… …
0.1 10
0.01 100
0.001 1000
0.0001 10000
… …

From the above table we can see that \lim\limits_{x \to 0-} f(x) = -\infty and \lim\limits_{x \to 0+} f(x) = +\infty

Clearly \lim\limits_{x \to 0-} f(x) \neq \lim\limits_{x \to 0+} f(x) i.e., left hand limit and right hand limit exist, but they are not equal.

Therefore \lim\limits_{x \to 0} \dfrac{1}{x} does not exists.

Question 4: Evaluate \lim\limits_{x \to -\infty} (1 + \dfrac{2}{x^2}).

Solution : Let f(x) = (1 + \dfrac{2}{x^2}).

x f(x)
-10 -1.02
-100 -1.0002
-1000 -1.000002
-10000 1.00000002
… …

From the table we can see that as x increases numerically without limit, assuming negative values, the numerical difference between f(x) and the finite value 1 can be made as small as we please. Hence, as x \to -\infty , f(x) \to 1.

Therefore \lim\limits_{x \to -\infty} f(x) = \lim\limits_{x \to -\infty} (1 + \dfrac{2}{x^2}) = 1.

Question 5: Evaluate

\lim\limits_{x \to 2} (2x^2 - 3x + 5)

\lim\limits_{x \to 3} \dfrac {x^2-9}{x-3}

Solutions:

a.

\lim\limits_{x \to 2} (2x^2 - 3x + 5)

= \lim\limits_{x \to 2} (2x^2) - \lim\limits_{x \to 2}3x + \lim\limits_{x \to 2}5

= 2 \lim\limits_{x \to 2} (x^2) - 3\lim\limits_{x \to 2}x + 5

= 2 (2^2) - 3\times2 + 5

= 7

b. Let f(x) = \dfrac {x^2-9}{x-3}

Now (x^2-9) = (x-3)(x+3)

Therefore f(x) = \lim\limits_{x \to 3} \dfrac {x^2-9}{x-3} = \lim\limits_{x \to 3} \dfrac {(x-3)(x+3)}{x-3}

= f(x) = \lim\limits_{x \to 3} (x+3)

= 3 + 3 = 6

Exercise

1. Show that f(x) = \lim\limits_{x \to -2} (\surd x^3 + 3x^2 - x + 3) = 3

2. Evaluate \lim\limits_{x \to 1} \dfrac{x^2-3x+2}{x^2-4x+3}. (Hint: factorize the numerator and the denominator)

3. Evaluate \lim\limits_{x \to 1} \dfrac{x^2-5x+6}{x^2-5x+2}.

4. Show that \lim\limits_{x \to -2} \dfrac{x^2-5x+3}{x^2+1}.

5. Evaluate \lim\limits_{x \to 2} \dfrac{x - \surd 3(x-2)}{x^2-4}. [Hint: Multiply both the numerator and the denominator by {x + \surd 3(x-2)}. and rationalize]

6. A function f(x) defined as follows:

f(x) = \begin{cases} 2x + 1, & \mbox{if } x\mbox{ \textless= 1 } \\ 3-x, & \mbox{if } x\mbox{ \textgreater 1} \end{cases}

Examine whether \lim\limits_{x \to 1} f(x) exists or not.

7. Evaluate \lim\limits_{x \to 0} \dfrac{e^{sin x}-1}{x} (Hint: Multiply both numerator and denominator by sin ⁡x. Put sin ⁡x = z)

8. Evaluate \lim\limits_{x \to 0} \dfrac{tan^2 x}{x^2} [Hint: Put tan^2 x = \dfrac{sin^2 x}{cos^2 x} ]

9. Evaluate \lim\limits_{x \to \pi} \dfrac {sin x}{\pi - x} [Hint: Put \pi - x = \theta ]

10. Evaluate \lim\limits_{x \to 2} \dfrac {x^{3/2} -2 \sqrt 2}{x-2}

Understanding Functions »


Filed Under: Calculus Tagged With: Limits

Comments

  1. shelly says

    September 4, 2015 at 4:21 am

    waooo i really love the way concept is given and elaborated with the help of solved examples..

    Reply
  2. upahar kc says

    December 26, 2021 at 1:40 pm

    oh really is it helpful for solving calcules question.

    Reply

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  • Understanding the Concept of Limits
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