Algebra of Limit

Table of Content

Can a function have two distinct limits as x approaches 'a'?
We can use several methods for evaluating the limits. Some of the important ones include
Some important theorems
Some Important Results on Limits
Frequently Used Series Expansions
Limits at infinity and infinite limits
Related Resources

We have already discussed the concept of limit of a function.

Let f be a function defined on an open interval containing a point ‘c’ (except possibly at c) and let us assume ‘L’ to be a real number.

Then, the function f is said to tend to a limit ‘L’ written as lim x→c f(x) = L if for each ∈ > 0, there exists a δ > 0 such that for all x satisfying 0 < |x – c| < δ, we have |f(x) – L| < ∈ or symbolically, we can define it as

∀ ∈ > 0, ∃ a δ > 0 such that 0 < |x – c| < δ ⇒ |f(x) – L| < ∈.

In this section, we shall mainly discuss the concept of algebra of limits and various important results

Suppose lim _x→a f(x) = α and lim _x→a g(x) = β then, if both α and β exist, then we can define the following rules:

lim _x→a [f(x) + g(x)] = lim _x→a f(x) + lim _x→a g(x) = α + β

lim _x→a [f(x) - g(x)] = lim _x→a f(x) + lim _x→a g(x) = α - β

lim _x→a [f(x) . g(x)] = lim _x→a f(x) + lim _x→a g(x) = α . β

$\lim_{x\rightarrow a}\left [ \frac{f(x)}{g(x)} \right ]=\frac{lim_{x\rightarrow a}f(x)}{lim_{x\rightarrow a}g(x)} = \frac{\alpha\ }{\beta }, provided \beta \neq 0$

lim _x→a fog(x) = f(lim _x→a g(x)) = f(β), only if f is continuous at g(x) = β

lim _x→a |f(x)| = |lim _x→a f(x)| = |α|

lim _x→a [f(x)]^g(x)= lim _x→a [f(x)]^lim^_x→a^g(x) = α^β

If lim _x→a f(x) = + ∞ or -∞ then

$\lim_{x\rightarrow a}\left [ \frac{1}{f(x)} \right ]=0$

If f(x) ≤ g(x) for every x in the neighborhood of a, then lim _x→a f(x) ≤ lim _x→a g(x).

For more on algebra of limits, view the following video

The above rules are applicable only when both lim _x_→_a f(x) and lim _x_→_a g(x) exist separately.

We now prove two of the properties, the rest of the parts can be easily proved as the proof of all the rules will follow on similar lines:

(i) lim _x_→_a (f(x) + g(x)) = lim _x_→_a f(x) + lim _x_→_a g(x)

To prove the above result we choose ∈, then by definition of limit, there exist δ₁ and δ₂ both greater than zero such that

algebra-of limits2

Let δ = min (δ₁,δ₂) then by algebra, we have

|f(x) + g(x) – α – β| = |f(x) – α + g(x) – β|

≤ |f(x) – α| + |g(x) - β|

< ∈ whenever 0 < |x - α| < δ

Hence lim _x→a (f(x) + g(x)) = lim _x→a f(x) + lim _x→a g(x)

(ii) lim _x_→_a (f(x).g(x)) = lim _x_→_a f(x) . lim _x_→_a g(x)

We have to prove that for ∀ ∈ > 0, there exists some δ > 0 such that |f(x)g(x) – αβ| < ∈ whenever 0 < |x-a| < δ

Now |f(x)g(x) - αβ| = |f(x)g(x) - αg(x) + αg(x) - αβ|

< |f(x)g(x) - αg(x)| + |αg(x) - αβ|

= ||f(x) - α| |g(x)| + |α| |g(x) - β|

From definition ∀ ∈ > 0, ∃ δ₁, δ₂ > 0 such that |f(x) – α | < ∈/3|β| whenever 0 < |x-a| < δ₁where β is the upper bound of g(x) in the neighborhood of x - a

and |g(x) – β| < ∈/2|α| whenever 0 < |x-α| < δ₂

So, we have

Now |f(x)g(x) - αβ| = ||f(x) - α| |g(x)| + |α| |g(x) - β|

< ∈ /3|β|. 3|β|/ 2 + |α| .∈ /2|α|

= ∈

So, |f(x)g(x) - αβ| < ∈ and hence lim _x→a (f(x).g(x)) = lim _x→a f(x) . lim _x→a g(x).

Illustration:

let f: R → R be such that f(1) = 3 and f’(1) = 6. The find the value of lim _x→0 [f(1 + x)/ f(1)]^1/x. (IITJEE 2002)

Solution:

Let y = [f(1 + x)/ f(1)]^1/x

So, log y = 1/x [log f(1+x) – log f(1)]

So, lim _x→0 log y = lim _x→0[1/f(1 + x) . f’(1 + x)]

= f’(1)/f(1)

= 6/3

log (lim _x→0y) = 2

lim _x→0 y = e²

Can a function have two distinct limits as x approaches 'a'?

If lim _x→a f(x) exists then it is unique i.e. there cannot be two distinct numbers L₁and L₂ such that when x → a the function tends to both L₁ and L₂. Hence if lim _x→a f(x) = L₁ and lim _x→a f(x) = L₂,then L₁= L₂.

If lim_x→af(x).g(x) exists then we can have the following cases:
Both the limits i.e. lim _x→a f(x) and lim _x→ag(x) exist.
lim _x→a f(x) exists while lim _x→ag(x) does not exist.

Consider f(x) = x and g(x) = 1/x, then lim _{x →0} f(x) g(x) exists and is equal to 1.

Also, lim _{x →0} f(x) exists and is equal to 0 but lim _{x →0} g(x) does not exist.

Both the limits i.e. lim _x→a f(x) and lim _x→ag(x) do not exist but even then lim _x→a f(x).g(x) exists.
Similarly, if lim_x→a [f(x) + g(x)] exists then we can have the following cases:
If ?lim _x→a f(x) exists then lim _x→ag(x) must exist.

This can be understood this way:

Since, g = (f + g) – f

So, by the limit theorem, lim _x→a g(x) = [lim _x→a f(x) + lim _x→a g(x)] - lim _x→a f(x).

Both the limits i.e. lim _x→a f(x) and lim _x→ag(x) do not exist.

Consider lim _x→1 [x] and lim _x→1 g{x}, where [.], {.} represent the greatest integer and fractional part function respectively. Here, both the limits do not exist but lim _x→1 [x] + {x} = lim _x→1 x = 1 exists.?

We can use several methods for evaluating the limits. Some of the important ones include

1. Direct Substitution Method:

Consider the following limits (i) lim _x→a f(x) and (ii) lim _x→a φ(x)/Ψ(x).

If f(a) and φ(a)/Ψ(a) exist and are fixed real numbers and Ψ(a) ≠ 0, then we say that lim _x→a f(x) = f(a) and lim _x→a φ(x)/Ψ(x) = φ(a)/Ψ(a).

?2. Factorization Method:

Consider lim _x→a f(x)/g(x). If by substituting x = a, f(x)/g(x) reduces to the form 0/0, then (x – a) is a factor of both f(x) and g(x). So, we first factorize f(x) and g(x) and then cancel out the common factor to evaluate the limit.

3. Rationalization Method:

This method is generally used in cases where either the numerator or the denominator or both involve expression consists of square roots and on substituting the value of x, the rational expression takes the form 0/0, ∞/∞.

4. Evaluation of Algebraic limits at infinity:

We know that lim_x→∞ 1/x = 0 and lim _x→∞ 1/x² = 0.

Then lim _x→∞ f(x) = lim _y→0 f(1/y).

Note:

When infinite limit is encountered i.e. '0' in the denominator as x→a, then either (x-a) gets cancelled from numerator and denominator both and if it is not possible then the limit's value is put as infinity. Such limits are called improper limits i.e. lim _x_→_∞ 1/x² = ∞.

Some important theorems

1. Let f < g in an open interval containing 'a' then lim _x→a f(x) < lim _x→ag(x).

2. Sandwich Theorem:

Let f, g, h be three continuous functions such that f < g < h in an open interval containing 'a' and suppose lim _x→a f(x) = lim _x→a h(x) = l. Then lim _x→a g(x) also exists and is equal to 'l'.

Illustration:

As shown in figure given below, we have f ≤ g ≤ h.

In (x₁, x₂) as x approaches a, lim _x→a f(x) = lim _x→a h(x) = lim _x→a g(x) = 1.

3. If 0 ≤ f ≤ g in an open interval containing a and lim _x→a g(x) = 0 then lim _x→a f(x) exists and is equal to zero.

4. lim _x→a f(x) = 0 if and only if lim _x→a f(x)

Some Important Results on Limits

Frequently Used Series Expansions

Following are some of the frequently used series expansions:

series expension

Illustration:

Using the formulae evaluate the following

$\lim_{x\rightarrow 0} \left ( \frac{sin 3x}{x} \right )$
$\lim_{x\rightarrow 0} \left \(1-x)^{1/x} \right$
lim _x→0 |x|/ x

Solution:

(1) The given function is

$\lim_{x\rightarrow 0} \left ( \frac{sin 3x}{x} \right )$

= lim _x→0 (3 sin x - 4 sin³x)/x

= lim _x→0 3(sin x /x) - lim _x→0 4(sin x /x) (sin x)²

= 3 × 1 – 4 × 1 × 0 = 3.

(2) The given function is

$\lim_{x\rightarrow 0} \left \(1-x)^{1/x} \right$ ?

= lim _x→0 (((1 + (- x))^1/-x)^-1

= 1/e

Note:

We know that

√(x²) = |x| = x ∀ x ≥ 0

= -x ∀ x < 0

lim _x→0⁺ |x|/x = lim _x→0⁺ x/x = lim _x→0⁺ 1 = 1

Also, lim _x→0^- |x|/x = lim _x→0^- (-x)/x = lim _x→0^--1 = -1

∴ lim _x→0⁺ |x|/x ≠ lim _x→0^- |-x|/x

∴ lim _x→0 |x|/x does not exist

Illustration:

Evaluate lim _x→2 (x²- 3x + 2)/(x²- x – 2).

Solution:

Let f(x) = (x²- 3x + 2)/(x²- x – 2)

First, let us assume that x→ 2⁺

i.e. x = 2 + h, h → 0, h > 0

lim _x→2⁺ f(x)

= lim _h→0 f(2+h)

= lim _h→0 ((2 + h)² - 3(2 + h) + 2)/((2 + h)² - (2 + h) - 2)

= lim _h→0 h(1 + h)/h(3 + h) (since h ≠ 0, we cancel h)

= lim _h→0 (1 + h)/(3 + h)

= (1 + 0)/(3 + 0) = 1/3

Now, consider the case when x → 2- i.e. x = 2 – h, h → 0, h > 0

lim _x→2^- f(x) = lim _h→0 f(2 – h)

= lim _h→0 ((2 – h)² - 3(2 – h) + 2)/((2 - h)²- (2 – h) – 2)

= lim_h→0 ((h²- h)/(h²- 3h))

= lim _h→0 (h – 1)/(h – 3)

= (0 – 1)/(0 – 3) = 1/3

∴ lim _x→2⁺ f(x) = lim _x→2^- f(x) = 1/3

Limits at infinity and infinite limits

When the term limits at infinity is used, it obviously means the evaluation of limit of functions as x→ ∞

For example lim _x→∞ 1/x² = 0

The term infinite limit means that when x tends to a particular value 'a'. Then the limit of the function tends to infinity i.e.lim _x→2 f(x) = ∞

Evaluation of the limit at infinity in a problem is solved by changing the expression f(x) to g(1/x) form. Here, we have to find the value of the function f(x) as x→∞.

Illustration:

lim _{x→ ∞} ((3x – 1)(4x – 2))/((x – 8)(x - 1))

Solution:

lim _{x→ ∞} ((3x – 1)(4x – 2))/((x – 8)(x - 1))

lim _{x→ ∞} ((3 – 1/x)(4 – 2/x))/((1 – 8/x)(1 – 1/x))

Given expression has been changed from f(x) to g(1/x) form.

_______________________________________________________________

Illustration:

For x ∈ R,

$\lim_{x\rightarrow \infty }[ \frac{x-3}{x+2}]^{x} = ?$ (IIT JEE 2000)

Solution:

$\lim_{x\rightarrow \infty }[ \frac{x-3}{x+2}]^{x} = \lim_{x\rightarrow \infty }\frac{(1-3/x)^{x}}{(1+2/x)^{x}}$

$=\frac{e^{-3}}{e^{-2}}$

= e^-5

Note:

When infinite limit is encountered i.e. '0' in the denominator as x→a, then either (x – a) get cancelled from numerator and denominator both and if it is not possible then the limit's value is put as infinity. Such limits are called improper limits i.e. lim_x→∞ 1/x² = ∞.

Q1. If lim _x_→_a f(x).g(x) exists then

(a) both the limits lim _x→a f(x) and lim _x→a g(x) must exist.

(b) atleast one of them must exist.

(d) none of these

Q2. If lim _x_→_a (f(x) + g(x)) exists then

(a) both the limits lim _x→a f(x) and lim _x→a g(x) must exist.

(b) if one exists then the other must exist

(d) none of these

Q3. lim _x_→₀ |x|/x

(a) does not exist.

(b) exists.

(d) none of these

Q4. lim _x_→_a f(g(x)) = f(lim _x_→_a (g(x)) = f(m)

(a) is always true

(b) true provided f is continuous at g(x) = m

(d) none of these

Q5. lim _x_→₀ (1+x)^1/x = ?

(a) 1

(b) 0

(d) ∞

Q1.	Q2.	Q3.	Q4.	Q5.
(d)	(b)	(a)	(b)	(c)

Related Resources

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