Limits and Derivatives Solutions

DEFINITION OF LIMIT

Let f(x) be a function of x. If for every positive number ε, however small it may be, there exists a positive number δ, such that whenever 0 < |x - a| < δ, we have f(x) - l| <ε, then we say ‘f(x) tends to the limit l as x tends to a’, and we write:

Lim (x → a) f(x) I.

Meaning of (x → a)

Let x be a variable and ‘a’ be a constant. x assumes values nearer and nearer to ‘a’, then we say ‘x tends to a’ and write ‘x → a’ and it doesn’t mean x = a.

Condition for Existence of Limit: Left and Right Limit

Let y = f(x) be a given function, and x = a is the point under consideration. Left tendency of f(x) at x = a is called its left limit and right tendency is called its right limit.
Left tendency (left limit) is denoted by f(a − 0) or f(a–) and right tendency (right limit) is denoted by f(a + 0) or f(a+) and are written as:

limits derivatives
where ‘h’ is a small positive number.

Thus for the existence of the limit of f(x) at x = a, it is necessary and sufficient that f(a - 0) = f(a + 0), if these are finite or f(a - 0) and f(a + 0) both should be either + ∞ or − ∞.

Frequently Used Limits

derivatives

limits and derivatives

Evaluation of Limits (Working Rules)

  1. By factorization: To evaluate limx→aφ(x)/ψ(x), factorise both φ(x) and ψ(x), if possible, then cancel the common factor involving ‘a’ from the numerator and the denominator. In the last, obtain the limit by substituting ‘x’ for ‘a’.
  2. By Substitution: To evaluate limx→a f(x), put x = a + h and simplify the numerator and denominator, then cancel the common factor involving h in the numerator and denominator. In the last obtain the limit by substituting h = 0.
  3. By L–Hospital’s Rule: Apply L–Hospital’s rule to the form 0/0 or ∞/∞.
  4. By Rationalisation: In case if numerator or denominator (or both) are irrational functions, rationalisation of numerator or denominator (or both) helps to obtain the limits of the function.
  5. In case of limits x → ∞, put x = 1/t or t = 1/x, so that t → 0 when x → ∞ and proceed as usual.
  6. To find the limit by finding L.H.L and R.H.L.: We evaluate the limit of a function f(x) at x → a by finding its right hand limit (R.H.L) and left hand limit (L.H.L) in the special case when the values of f(x) are given by different functions of x for x < a and for x ≥ a.
(vii) Methods for calculating the limits of the form limx→a (f(x))g(x)

Here we discuss two different cases,

(a) When limx→a f(x) = 1:
In this case,
limx→a [f(x)]g(x) = limx→a [1 + f(x) - 1]g(x)
= limx→a ( [1 + f(x) - 1]1/(f(x)-1) )g(x)(f(x)-1)
= elimx→a g(x)(f(x)-1)
because, limx→a ( [1 + f(x) - 1]1/(f(x)-1) ) = e

(b) When limx→a f(x) ≠ 1 but f(x) is positive in the neighbourhood of x = a.
In this case we write,
(f(x))g(x) = eg(x)·ln(f(x))
⇒ limx→a [f(x)]g(x) = elimx→a g(x)·ln(f(x))

Remark:

  • If f(x) is not throughout positive in the neighbourhood of x = a, then limx→a (f(x))g(x) will not exist. Because in this case function will not be defined in the neighbourhood of x = a.

L'Hospital’s Rule

Let f(x) and g(x) be two functions differentiable in the neighbourhood of the point a, except at the point ‘a’ itself. If limx→a f(x) = limx→a g(x) = 0. Or, limx→a f(x) = limx→a g(x) = ∞. Then limx→a [f(x)/g(x)] = limx→a [f'(x)/g'(x)] provided that the limit on the right exist as a finite number or is ±∞.

Remark:
  • L’Hospitals rule is not always useful. Consider the example, limx → ∞ (x + sin x) / (x - sin x) (form ∞/∞).

Here, if we apply L’Hospital’s rule, then limx → ∞ (x + sin x) / (x - sin x) = limx → ∞ (1 + cos x) / (1 - cos x) .

Now, both the numerator and denominator are undefined because limx → ∞ cos x doesn’t exist.

We can find this limit as follows:

limx → ∞ (x + sin x) / (x - sin x) = limx → ∞ (1 + (sin x / x)) / (1 - (sin x / x)) = (1 + 0) / (1 - 0) = 1   since limx → ∞ (sin x / x) = 0 .

Use of Sandwich Theorem in Solving Problems

Sandwich theorem helps in calculating the limits, when limits can not be calculated using the usual methods. The following illustration would make the procedure clear,

CONTINUITY OF A FUNCTION
A function f(x) is said to be continuous at x = a if
  limx→a- f(x) = limx→a+ f(x) = f(a)

If f(x) is not continuous at x = a, we say that f(x) is discontinuous at x = a.

f(x) will be discontinuous at x = a in any of the following cases:
(i)  limx→a- f(x) and limx→a+ f(x) exist but are not equal.
(ii)  limx→a- f(x) and limx→a+ f(x) exist and are equal but not equal to f(a).
(iii) f(a) is not defined.
(iv) At least one of the limits does not exist.

limits and countinity

DIFFERENTIABILITY

Let y = f(x) be continuous in (a, b). Then the derivative or differential of f(x) at x ∈ (a, b), denoted by dy/dx or f′(x), and is defined as

dy/dx definition
provided the limit exists and is finite.

Right hand derivative

Right hand derivative of f(x) at x = a is denoted by, Rf′(a) or f′(a+) and defined as

Right hand derivative

Left hand derivative

Left hand derivative of f(x) at x = a is denoted by Lf′(a) or f′(a−) and is defined as

Left hand derivative

Clearly, f(x) is differentiable at x = a if and only if Rf′(a) = Lf′(a).

Note:
If a function f(x) is differentiable at x = a then it is also continuous at x = a. But if a function is continuous at a point, it is not necessarily differentiable at that point.

Formulas and Concepts

  1. For the existence of the limit at x = a, f(x) need not be defined at x = a. However if f(a) exists, limit need not exist or even if it exists then it need not be equal to f(a). limx→a f(g(x)) = f(limx→a g(x)) = f(ℓ2) , if and only if f(x) is continuous at x = ℓ2.
  2. Right hand derivative of f(x) at x = a is denoted by, Rf'(a) or f'(a+) and is defined as
    R f'(a) = limh→0 [f(a+h) – f(a)] / h, h > 0.
  3. Left hand derivative of f(x) at x = a is denoted by Lf'(a) or f'(a–) and is defined as
    L f'(a) = limh→0 [f(a–h) – f(a)] / –h, h > 0.
  4. f(x) is differentiable at x = a if and only if R f'(a) = L f'(a).
  5. L' Hospital's Rule

We have dealt with problems which had indeterminate form either 0/0 or ∞/∞.
The other indeterminate forms are ∞ − ∞, 0·∞, 00, ∞0, 1∞.
We state below a rule, called L' Hospital's Rule, meant for problems on limit of the form 0/0 or ∞/∞.

Let f(x) and g(x) be functions differentiable in the neighbourhood of the point a, except may be at the point a itself. If

limx→a f(x) = 0 = limx→a g(x)
or,
limx→a f(x) = ∞ = limx→a g(x),
then
limx→af(x)/g(x) = limx→af'(x)/g'(x)
provided that the limit on the right either exists as a finite number or is ±∞.

  1. If limx → a f(x) = 0, then the following results will be holding true:
    1. limx → a ¼sin f(x)⁄f(x)¾ = limx → a ¼tan f(x)⁄f(x)¾ = limx → a cos f(x) = 1
    2. limx → a ¼sin−1 f(x)⁄f(x)¾ = limx → a ¼tan−1 f(x)⁄f(x)¾ = 1
    3. limx → a ¼bf(x) - 1⁄f(x)¾ = ln b (b > 0)
    4. limx → a (1 + f(x))1/f(x) = e

SOLVED EXAMPLES

1.    equals 
         (A) 2          (B) 0 
         (C) –2       (D) none of these 

Sol.    (D).   . 
                 
    LHL ≠ RHL 
    Limit does not exist. 

2.    In order that the function f(x) = (x +1)cotx is continuous at x = 0, f(0) must be defined as 
    (A) f(0) = 0    (B)  f(0) = e 
    (C)  f(0) = 1/ e     (D) none of these 

Sol.    (B). For continuity actual value must be equal to limiting value
    A = limx→0(x+1)cotx 
    logA = limx→0cotxlog(x+1) 
    = limx→0log(x+1)/tanx                [0/0 form]
    =        (By L’ Hospital Rule) 
    log A = 1  ⇒ A = e1 = e . 
    For  f(0) must be defined as f(0) = e . 

3.    If f(x) = . Then
    (A)  f is not a continuous function    
    (B) f' (0+) exists but  f' (0–) does not exist
    (C) f' (0+) = f' (0–)    
    (D) f' (0+) and f' (0–) does not exist

Sol.    (D). We have limx→0f(x)= limx→0xsin1/x = 0 = f(0) So, f is continuous at x = 0. 
    Therefore, f is a continuous function. As for the derivative f’(0)  we have.
    f' (0+) =   =   which doesn’t exist 
    Similarly, the limit f' (0–) doesn’t exist.

4.    is equal to
              (A) 2               (B) 3
              (C) 4               (D) 5
Sol.    (A).  = 2

5.    If f (x) = cos{π/2[x]-x3}, 1 < x < 2 (where [.] denotes the greatest integer function) then its derivative at x = (π/2)1/3  is equal to  
            (A) 0                    (B) 3(π/2)2/3 
           (C) –3(π/2)3/2     (D) none of these 

Sol.    (A). x = ((π/2)1/3 ⇒ [x] = 1 ; So, f (x) = cos(π/2-x3)= sin x3; f'(x) = 3x2 cos x3
    f'((π/2)1/3)= 3(π/3)2/3  cosπ/2 = 0

6.    If f (9) = 9, f' (9) = 4, then   is equal to
           (A) 9              (B) 4
         (C) 3               (D) 2

Sol.    (B). f (9) = 9, f' (9) = 4 
               

7.    Let f : R ⟶ R be a  function defined by f(x) = max{x, x3}. The  set  of all points where f(x) is NOT differentiable is 
    (A) {–1, 1}     (B) {–1, 0) 
    (C) {0, 1}     (D) {–1, 0, 1}
Sol.    (D). f (x) = max (x, x3)
    y = x, y = x3 
    x3 = x 
    x = 0, x = 1, – 1 
    from figure, it is very clear that 
    x is not differentiable at x = 0, – 1, 1 
     
8.    If f(x) =(x2+5x+3/x2+x+2)x, then limx→∞f(x) is 
    (A) e4    (B) e3
    (C) e2    (D) 24 

Sol.    (A).  = e4.

9.    If   = –1, then 
            (A) a = 1         (B) a = 0 
           (C) a = e         (D) none of these 

Sol.    (A). Applying L’Hospital rule
         axIna - axa-1/xx(1+In x)=-1 ⇒ ln a – 1 = –(1 + ln a) ⇒ ln a = 0 ⇒ a = 1.

10.    The value of  limx→0(cosx + asinbx)1/x is 
         (A) 1              (B) ab
        (C) eab           (D) eb/a

Sol.    (C).  =  
 

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