Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Limit of functions

The limit of a function is the most fundamental concept in the study of Calculus as well as of mathematical analysis.

Limit of a function  f(x) at a point  a is defined by  \lim_{x \to a} f(x)=f(a) where  x\longrightarrow a is read as  x  approaches (or tends) to  a and   x \neq a.

Consider a function  f defined by  f(x)=2x+1 .

The domain of this function is the set of real numbers.  1 \in D_f .

It is natural to ask whether or not  2x+1 becomes closer and closer to   3 as  x  becomes closer and closer to   1  but not equal to  1 .

Hence  \lim_{x \to 1} (2x+1)=3 i.e.,  2x+1\longrightarrow 3 as  x\longrightarrow 1

The statement is also expressed by saying that limit of  (2x+1)  at  x=1 is equal to  3

Algebra of Limits:

Let  f and  g be two functions such that both  \lim_{x \to a} f(x) and  \lim_{x \to a} g(x) exists , then

  •  \lim_{x \to a} \left[ f(x)+g(x) \right]=\lim_{x \to a} f(x)+\lim_{x \to a} g(x)
  •  \lim_{x \to a} \left[ f(x)-g(x) \right]=\lim_{x \to a} f(x)-\lim_{x \to a} g(x)
  •  \lim_{x \to a} \left[ f(x) \cdot g(x) \right]=\lim_{x \to a} f(x)\cdot \lim_{x \to a} g(x)
  •  \lim_{x \to a} \dfrac{f(x)}{g(x)}=\dfrac{\lim_{x \to a f(x)} }{\lim_{x \to a}g(x) }
  • If  g be a constant function such that g(x) =\lambda , for some real number  \lambda .  \lim_{x \to a} \left[ (\lambda \cdot f)(x) \right]=\lambda\lim_{x \to a} f(x)

Example 1:

Find  \lim_{x \to 1} ~[x^3 - x^2 + 1]

Solution:

 \lim_{x \to 1}~ x^3 - x^2 + 1 = 1^3 - 1^2 + 1 = 1

Example 2:

Find  \lim_{x \to 3} [x(x + 1)]

Solution:

 \lim_{x \to 3}~ [x (x + 1)] = 3(3 + 1) = 12

Example 3:

 \lim_{x \to 1} \left[ \dfrac{x^2 + 1}{x + 100} \right] = \lim_{x \to 1} \left[\dfrac{1^2 + 1}{1 + 100} \right] = \dfrac{2}{101}

A point  x  is at a distance less than  \delta (~\delta > 0) from the point  a , then the set of all  x \neq a such that  x is at a distance less than any fixed  \delta > 0 from  a is expressed as

 x is arbitrarily close to  a and  x \neq a ”.

The neighborhood of a Point :

If  a\in R , then any open interval containing the point  a is called a neighborhood of a point.

The open interval  (a - \delta, ~a + \delta) is called the  \delta – neighborhood of the point  a where  \delta > 0 .

\begin{aligned}\text {(i)} \text { The set }(a - \delta, a + \delta)&= \left\{ x \in R |~ a - \delta < x < a + \delta \right\} \\&= \left\{ x \in R |~ |x - a| < \delta \right\} \end{aligned}

 \begin{aligned} \text {(ii)} (a- \delta, a + \delta) - \left\{ \operatorname{a} \right\}& = \left\{ x \in R~| a - \delta < x , a + \delta \text { & } x \neq a \right\} \\&=  \left\{ x \in R ~|~ 0 < |x - a| < \delta \right\} \end{aligned}

(iii)  (a - \delta, a) = \left\{ x \in R ~|~ a - \delta < x < a \right\}

(iv) (a, a + \delta) = \left\{ x \in R ~|~ a < x < a + \delta \right\}

Definition of Limit of Function:

A number  l is called the limit of a function  f as  x tends to  a i.e. \lim_{x \to a} f(x) = l if for any  \in > 0 there exist  \delta > 0 depending on  \in such that :

 0 < |x - a| < \delta

 \implies |f(x) - l| < \in

i.e. a - \delta < x < a + \delta and  x \neq a

 \implies l - \in < f(x) < l + \in}

Example:

 \lim_{x \to a} x = a

Solution:

Let  \in > 0 , we take  f(x) = x then  |f(x) - a| < \in if  |x - a| < \in taking  \delta = \in we see that there exists  (\exists) \delta > 0  depending in  \in such that:

 |x - a| < \delta \implies |f(x) - a| < \in

So,  0< |x - a| < \delta \implies |f(x) - a| < \in

Hence  \lim_{x \to a} x = a

i.e., \lim_{x \to a} x=a

Example 1:

Solve:  \lim_{x \to 1} \left\{  {\sqrt{x} + x + \dfrac{1}{\sqrt{x}} \right\}

Solution:

 \lim_{x \to 1} \left\{ \sqrt{x} + x + \dfrac{1}{\sqrt{x}} \right\}

 \lim_{x \to 1} \sqrt{x} + \lim_{x \to 1} x + \lim_{x \to 1} \dfrac{1}{\sqrt{x}}

 = 1 + 1 + 1 = 3

Example 2:

 \lim_{x \to 0} (2x + 1) (\sqrt{x} + 5)

 = \lim_{x \to 0} (2x + 1) \cdot \lim_{x \to 0} (\sqrt{x} + 5) = 1 \times 5 = 5

Example 3:

 \lim_{x \to 1} \left\(\dfrac{x + \sqrt{x}}{2x + 1} \right\) = \dfrac{\lim_{x \to 1} (x + \sqrt{x})}{\lim_{x \to 1} (2x + 1)} = \dfrac{1 + 1}{2 + 1}=\dfrac{2}{3}

Left hand and Right-hand Limit of a Function

For any  \in > 0 there exists   ( \exists ) \delta > 0 depending on  \in such that  a - \delta < x < a + \delta and  x \neq a \implies l - \in < f(x) < l + \in , then  \lim_{x \to a} f(x) = l

 a - \delta < x < a + \delta and  x \neq a means  a - \delta < x < a or  a - x < a + \delta

So if (i)  a - \delta < x < a \implies l - \in < f(x) < l + \in and (ii)  a < x < a + \delta \implies l - \in < f(x) < l + \in

then  \lim_{x \to a} f(x) = l

Left hand limit

A number  l_1 is called the left-hand limit of  f{x} at  x = a or simply  \lim_{x \to a^-} f(x) = l_1 if for any  \in > 0 there exists  \delta > 0 depending on  \in such that  a - \delta < x < a  \implies |f(x) - l_1| < \in

Right-hand limit:

A number  l_2 is called the right-hand limit of  f(x) at  x = a it for any  \in > 0 there exists  \delta > 0 depending in  \in such that  a < x < a + \delta \implies |f(x) - l_2| < \in i.e.  \lim_{x \to a^+} f(x) = l_2

Note-:

(i) If  l_1 = l_2 = l

i.e.  \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = l , then  \lim_{x \to a} f(x) exists and equal to  l

i.e.  \lim_{x \to a} f(x) = l

(ii) If  l_1 \neq l_2 , then  \lim_{x \to a} f(x) does not exist.

(iii) For a function  f to have a limit at a point  a \in R it is necessary and sufficient that both  \lim_{x \to a^-} f(x) and  \lim_{x \to a^+} f(x) exist, and these limits coincide.

Infinite limits

(i)  \lim_{x \to a} f(x) = \infty ,if given  G > 0 , there exists  \delta > 0 depending on  G such that   0 < |x-a| < \delta \implies f(x) > G .

(ii)  \lim_{x \to a} f(x) = \infty , if given  G > 0 there exists  \delta > 0 depending on  G such that   0 < |x - a| < \delta \implies f(x) < - G where  'G' is as large as possible number greater than  0.

Limits at Infinity:

(i)  \lim_{x \to \infty} f(x) = l if given  \in > 0 , there exists  G > 0 depending on  \in such that  x > G \implies |f(x) - l| < \in

(ii)  \lim_{x \to \infty} f(x) = l if given  \in > 0 , there exists  G > 0 depending on  \in such that x < - G \implies |f(x) - l| < \in

(iii)  \lim_{x \to \infty} f(x) = \infty if given  G > 0 , there exists  K > 0 such that  x > K \implies f(x) > G .The concepts  \lim_{x \to - \infty} f(x) = -\infty and \lim_{x \to -\infty} f(x)=\infty are also defined by infinite limits at infinity.

Example:

Find  \lim_{x \to \infty} \dfrac{1}{x}

Solution:

We choose  \in > 0 , then  \left| \dfrac{1}{x} - 0  \right| = \dfrac{1}{x} = \dfrac{1}{x} < \in if  x > \dfrac{1}{\in} (\therefore x > 0) we take  G = \dfrac{1}{\in} .

Thus given  \in > 0 , there exists  G > 0 depending on  \in such that  x > G \implies \left| \dfrac{1}{x} -0 \right| < \in .

So  \dfrac{1}{x} \to 0 as  x \to \infty

i.e. \lim_{x \to \infty} \dfrac{1}{x} = 0

Note:

If  x increases indefinitely through positive values,  \dfrac{1}{x} remains positive and decreases indefinitely. So intuitively we see that  \dfrac{1}{x} \to 0 as  x \to \infty .

Note:

 \lim_{x \to a} f(x) = l and  \lim_{x \to a} g(x) = m can be written as  \lim_{x \to \infty} f(x) = l and in that case  \lim_{x \to - \infty} f(x) = m , holds if  1a is  x \to - \infty replaced by  \infty and in second case  a is replaced by  - \infty

Theorem:

(i) If a function  f satisfies the inequality   f(x) > M > 0 in a neighborhood of  a and a function  g is such that  \lim_{x \to a} g(x) = 0   g(x)\neq 0   for   x \neq a   and  g(x) > 0, \lim_{x \to a} \dfrac{f(x)}{f(x)} = \infty .

(ii) If  f(x) > M > 0 in a neighborhood of the point  a and if  g is such that  \lim_{x \to a}g(x) 0 ,  g(x) \neq 0   for   x \neq a   and  then  g(x) < 0,\lim_{x \to a} \dfrac{f(x)}{g(x)} = - \infty

Sandwich theorem / Squeezing theorem:

If \lim_{x \to a} f(x)-\lim_{x \to a} g(x)-l and a function  \phi is such that f(x) \leq ~\phi (x)~ \leq~ g(x) for all  x in a deleted neighborhood of  a , then \lim_{x \to a} \phi (x) =l .

Example:

If  \phi (x)=\begin{Bmatrix}x \sin \dfrac{1}{x}& (x \neq 0) \\0 & (x=0) \\\end{Bmatrix}, then find \lim_{x \to 0} \phi (x) by sandwich theorem.

Solution:

Since \left| \sin \dfrac{1}{x} \right| \leq 1 , \implies 0 \leq |\phi (x)|=|x \sin \dfrac{1}{x}|\leq |x| with  f(x)=0,  g(x)=|x| we have  \lim_{x \to 0} f(x)=0,~\lim_{x \to 0} g(x) and so by Sandwich Theorem:

 \lim_{x \to 0} \phi (x)=0

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