Limits, Continuity and Differentiability notes PDF

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The notes contain lectures of class I. It has all limit properties, formulas and various solved examples....

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Subject:- Mathematics Topic:- Limits, Continuity and Differentiability Class:-I Level:- 10+2 and all entrances Limit of a Fun Function:ction:Let 𝑦 = 𝑓(𝑥) be a function of 𝑥 . If at 𝑥 = 𝑎, 𝑓(𝑥) takes indeterminate form, then we consider the values of the function which are very close to 𝑎. If these values tend to a definite unique number as 𝑥 tends to 𝑎, then the unique number so obtained is called the limit of 𝑓(𝑥) at 𝑥 = 𝑎 and we can write it as lim 𝑓(𝑥). 𝑥→𝑎

Right Hand Limit ::-If 𝑥 approaches from the right, i.e. from larger values of 𝑥 than 𝑎, then the limit of 𝑓(𝑥) is called the Right Hand Limit (RHL) and is written as, lim 𝑓(𝑥) or

𝑥→𝑎+

lim 𝑓(𝑥)𝑜𝑟 𝑓(𝑎 + 0)

𝑥→(𝑎+0)

To find RHL, put 𝑥 = 𝑎 + 𝑏 and replace 𝑥 → 𝑎+ by 𝑏 → 0 and then simplifying lim 𝑓(𝑎 + 𝑏) 𝑏→0

using appropriate formula.

Left Hand Limit :If 𝑥 approaches 𝑎 from the left, i.e. from the smaller values of 𝑥 than 𝑎, then the limit of 𝑓(𝑥) is called the Left Hand Limit(LHL) and is written as, lim− 𝑓(𝑥) or 𝑓(𝑎 − 0). 𝑥→𝑎

To find LHL, put 𝑥 = 𝑎 − 𝑏 and replace 𝑥 → 𝑎− by b →0 and then simplify lim 𝑓(𝑎 − 𝑏) 𝑏→0

using appropriate formula. When the left hand limit equals to right hand limit, we say that

the function has a limiting values. Thus, for the existence of lim 𝑓(𝑥), the necessary and sufficient condition is lim 𝑓(𝑥) = lim 𝑓(𝑥) 𝑜𝑟 𝑓(𝑎 + 𝑥→𝑎 𝑏) = 𝑓(𝑎 − 𝑏) 𝑥→𝑎−

𝑥→𝑎+

Important Results Related to L Limit imit ::-i. ii. iii. iv. v. vi. vii. viii. ix.

lim[𝑓(𝑥) ± ∅(𝑥)] = lim 𝑓(𝑥) ± lim ∅(𝑥)

𝑥→𝑎

𝑥→𝑎

𝑥→𝑎

lim[𝑐. 𝑓(𝑥)] = 𝑐 lim 𝑓(𝑥), where c is a constant

𝑥→𝑎

𝑥→𝑎

lim[𝑓(𝑥). ∅(𝑥)] = lim 𝑓(𝑥) . lim ∅(𝑥)

𝑥→𝑎

lim

𝑥→𝑎

𝑓(𝑥)

= ∅(𝑥)

[lim 𝑓(𝑥)] 𝑥→𝑎

[lim ∅(𝑥)] 𝑥→𝑎

𝑥→𝑎

𝑥→𝑎

, provided lim ∅(𝑥) ≠ 0 𝑥→𝑎

lim 𝑙𝑜𝑔𝑓(𝑥) = log [lim 𝑓(𝑥)], provided lim 𝑓(𝑥) > 0 𝑥→𝑎

lim 𝑒

𝑥→𝑎

𝑓(𝑥)

𝑥→𝑎 lim 𝑓(𝑥)

= 𝑒 𝑥→𝑎

1

lim

𝑥→𝑎

𝑓(𝑥)

lim[ 1 + 𝑓(𝑥)]∅(𝑥) = 𝑒 𝑥→𝑎 ∅(𝑥)

𝑥→𝑎

lim ∅(𝑥)

lim[ 𝑓(𝑥)]∅(𝑥) = {lim 𝑓(𝑥)}𝑥→𝑎 𝑥→𝑎

𝑥→𝑎

If lim 𝑓[∅(𝑥)] = 𝑓[lim ∅(𝑥)], provided ‘𝑓’ is continuous at ∅(𝑥) ∈ ℝ 𝑥→𝑎

𝑥→𝑎

Example 1. If 𝑓(𝑥) is defined as 𝑓(𝑥) = { lim𝑥→1 𝑓(𝑥) exist?

2𝑘𝑥 + 3, 𝑖𝑓 𝑥 < 1 then for what values of k does 1 − 𝑘𝑥², 𝑖𝑓 𝑥 > 1

A. We have, lim− 𝑓(𝑥) = lim 𝑓(1 − ℎ) = lim[2𝑘(1 − ℎ) + 3] 𝑥→1

ℎ→0

ℎ→0

= [2𝑘(1 − 0) + 3] = 2𝑘 + 3

and lim+ 𝑓(𝑥) = lim 𝑓(1 + ℎ) = lim[1 − 𝑘 (1 + ℎ)2 ] 𝑥→1

ℎ→0

ℎ→0

= [1 − 𝑘(1 + 0)2 ] = 1 − 𝑘

Now, lim 𝑓(𝑥) exists, if lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) 𝑥→1

⇨ 2𝑘 + 3 = 1 − 𝑘

𝑥→1

⇨ 3𝑘 = −2

𝑥→1

⇨𝑘=−

3

2

Sandwich Theorem (or Squeeze theorem theorem):):-

If 𝑓, 𝑔 and 𝑏 are functions such that 𝑓(𝑥) ≤ 𝑔(𝑥) ≤ 𝑏(𝑥) for all 𝑥 in some neighbourhood of the point 𝑎 (except possibly at 𝑥 = 𝑎) and if lim 𝑓(𝑥) = 𝑙 = lim 𝑏(𝑥), then lim 𝑓(𝑥) = 𝑙 . 𝑥→𝑎

𝑥+7𝑠𝑖𝑛𝑥 𝑥→∞ −2𝑥+13

Example 2. Find the value of lim

𝑥→𝑎

𝑥→𝑎

, using sandwich theorem?

A. We know that, −1 ≤ 𝑠𝑖𝑛𝑥 ≤ 1 for all 𝑥

⇨ −7 ≤ 7𝑠𝑖𝑛𝑥 ≤ 7 ⇨ 𝑥 − 7 ≤ 𝑥 + 7𝑠𝑖𝑛𝑥 ≤ 𝑥 + 7

Now, dividing throughout by (−2𝑥 + 13), we get 𝑥−7

≥ −2𝑥+3

𝑥+7𝑠𝑖𝑛𝑥

−2𝑥+13

Now, lim

𝑥→∞

lim

≥

𝑥+7

−2𝑥+13

𝑥−7

= lim −2𝑥+13

𝑥+7

𝑥→∞ −2𝑥+13

𝑥→∞

= lim

𝑥→∞

7 𝑥 13 −2+ 𝑥

1+

for all 𝑥 that are large. 7

1− 𝑥

13 −2+ 𝑥

=

=

1+0 −2+0

1−0

1

=−2 −2+0 1

= −2

Therefore, by sandwich theorem, 𝑥+7𝑠𝑖𝑛𝑥 𝑥→∞ −2𝑥+13

lim

=−

1

2

Limit of a Rational Function: Function:-Limit of a rational function 𝑓(𝑥), of the form lim substitution methods.

𝑝(𝑥)

𝑥→𝑎 𝑞(𝑥)

, can be find out using factorization or

Indeterminate Form:If the expression obtained after substitution of value of the limit give the following forms 0

00 , , 1∞ , ∞ − ∞, 0

∞

∞

, 0 × ∞ and ∞0 , then it is known as an indeterminate form.

Example 3. If lim A. We have, ⇨ lim [ ⇨

𝑥→1

32 32 𝑥 4 −1 = lim 𝑥 −𝑘 𝑥→1 𝑥−1 𝑥→𝑘 4 −1 3 3 𝑥 lim = lim 𝑥𝑥 2−𝑘 −𝑘 2 𝑥−1 𝑥→1 𝑥→𝑘

(𝑥 2 +1)(𝑥+1)(𝑥−1) 𝑥−1

, then find the value of k?

(𝑥−𝑘)(𝑥 2 +𝑥𝑘+𝑘2 ) (𝑥−𝑘)(𝑥+𝑘) 𝑥→𝑘

] = lim [

]

𝑥 2 +𝑥𝑘+𝑘2 𝑥+𝑘 𝑥→𝑘

lim (𝑥 2 + 1)(𝑥 + 1) = lim 𝑥→1

(12 + 1)(1 + 1) =

⇨

⇨4=

3𝑘 2

𝑘 2 +𝑘.𝑘+𝑘 2

⇨ 3𝑘² − 8𝑘 = 0

2𝑘

𝑘(3𝑘 − 8) = 0

⇨

𝑘+𝑘

⇨ 𝑘 =0,

8

3

Since, 𝑘 = 0 does not saisfy the given equation, therefore 𝑘 =

8

3

Exponential and Logarithm Logarithmic ic Limits :For finding the limits of exponential and logarithmic functions, following results are useful 𝑎𝑥 −1 𝑥→0 𝑥 𝑒 𝑥−1 lim 𝑥 𝑥→0

= log𝑒 𝑎 , 𝑎 > 0

1. lim 2.

=1 1

1 𝑛

3. lim(1 + 𝑥)𝑥 = lim (1 + ) = 𝑒 𝑛 𝑥→0

4. lim(1 + 𝑥→0

5. lim

log 𝑥

𝑎 𝑥 ) 𝑥

𝑥→0

7. lim 𝑥→0

𝑥

log(1+𝑥) 𝑥

=𝑒

= 0, (𝑚 > 0)

𝑥→0 𝑥 𝑚 log𝑒 (1+𝑥)

6. lim

𝑛→∞ 𝑎

= log 𝑒, (𝑎 > 0, 𝑎 ≠ 1)

=1

𝑎

8. lim(1 + 𝜆𝑥)1/𝑥 = 𝑒 𝜆 𝑥→0

Example 4. Find the value of lim

𝑥→0

log(5+𝑥)−log(5−𝑥) 𝑥

?

A. We have, lim log(5+𝑥)−log(5−𝑥) =

𝑥 𝑥→0 𝑥 )} )}−log{5(1− 5 𝑥5 lim log{5(1+ 𝑥 𝑥→0

= lim

𝑥→0

= lim

𝑥→0

= lim

𝑥→0

𝑥

0 … … … … … … . [ 0 form]

𝑥

{log 5+log(1+ )}−{log 5+log(1− 5)} 5

𝑥

x

x

log(1+ 5)−log(1− 5) 𝑥 5

𝑥

log(1+ ) 𝑥/5

1 .5

+ lim

𝑥→0

x 5

log(1− ) −x/5

1

1

. (− 5) = 5 +

Trigonometric Limits ::-(i) (ii) (iii) (iv) (v) (vi)

lim

𝑥→0

𝑠𝑖𝑛𝑥 𝑥

=1

lim sin 𝑥 = 0

𝑥→0

lim cos 𝑥 = 1

𝑥→0

lim

𝑥→0

𝑡𝑎𝑛𝑥

=1

𝑥

𝑠𝑖𝑛−1 𝑥 𝑥 𝑥→0

lim

𝑡𝑎𝑛−1 𝑥 𝑥 𝑥→0

lim

=1

=1

Example 5. Find the value of lim

𝑥→0

𝑠𝑖𝑛|𝑥| 𝑥

sin(−𝑥) 𝑠𝑖𝑛|𝑥| , LHL= lim − 𝑥 𝑥→0 𝑥→0 𝑥

A. Since lim = − lim 𝑥→0

𝑠𝑖𝑛𝑥 𝑥

= −1

and RHL= lim+ 𝑥→0

𝑠𝑖𝑛𝑥

since RHL≠LHL

𝑥

=1

So, the limit does not exist.

?

1

2

=5 5

𝜋 Example 6. Evaluate lim 𝑥→

A. lim𝜋 𝑥→

2

= lim𝜋

2

1 − (𝑠𝑒𝑐𝑥 − 𝑡𝑎𝑛𝑥) = 𝑥→ lim𝜋 (𝑐𝑜𝑠𝑥 1−𝑠𝑖𝑛𝑥

𝑥→ 2 𝑐𝑜𝑠𝑥 𝑥

=

2

2

𝑠𝑖𝑛𝑥 ) 𝑐𝑜𝑠𝑥

𝑥 2

𝑥 (cos −sin 2) lim𝜋 [ 2𝑥2 𝑥 ] cos −sin2 2 𝑥→

𝑥

cos −sin lim𝜋 𝑥2 𝑥2 𝑥→ cos +sin 2

=

(𝑠𝑒𝑐𝑥 − 𝑡𝑎𝑛𝑥) .

2

2

=

2

𝜋

𝜋

cos −sin 4 4 𝜋 𝜋 cos +sin 4 4

=

1 1 − √2 √2 1 1 + √2 √2

=0

L’ HOSPITAL’S RULE ::-In this method, we first check , whether the form of the function after substituting the limit 0

is or not. 0 If it is not of this form, then make necessary operation in the function otherwise we differentiate both numerator and denominator with respect to 𝑥 . Differentiation can be done 𝑛 number of times according to the problem. The above rule can be applied for other indeterminate forms such as ∞

∞

, ∞ − ∞, 0 × ∞, 1∞ , 00 𝑎𝑛𝑑 ∞0 etc. 𝑠𝑖𝑛−1 𝑥−𝑡𝑎𝑛−1 𝑥 𝑥³ 𝑥→0

Example 7. Find the value of lim A. We have, lim

𝑥→0

= lim

𝑥→0

𝑠𝑖𝑛−1 𝑥−𝑡𝑎𝑛−1 𝑥 𝑥³

[(1+𝑥2 )²−(1−𝑥 2 )]

[3𝑥 2 √1−𝑥2 (1+𝑥2 )]

𝑥 4 +3𝑥 2 𝑥→0 3𝑥2 √1−𝑥2 (1+𝑥2 )

= lim

𝑥 2 +3

×

×

= lim 3(√1−𝑥2 )(1+𝑥 2) × 𝑥→0

= lim 1

[(1+𝑥2 )−√1+𝑥2 ]

𝑥→0 [3𝑥 2 √1−𝑥2 (1+𝑥 2 )]

(1+𝑥 2 )(√1−𝑥2 ) 1

(1+𝑥 2 )+√1−𝑥2 1

?

1+𝑥2 +√1−𝑥 2

= 1/2

[using L’ Hospital Rule, as

[𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑠𝑒]

0

0

form]...