USE L HOPITAL S RULE TO EVALUATE THE LIMIT

Evaluating Limits :

While computing the limits

lim x ->a R(x)

of certain functions R(x) , we may come across the following situations like,

The l’Hopital’s Rule :

Suppose f (x) and g(x) are differentiable functions and g'(x) ≠ 0 with

We say that they have the form of a number. But values cannot be assigned to them in a way that is consistent with the usual rules of addition and multiplication of numbers.

We call these expressions Indeterminate forms.

Evaluate the following limits, if necessary use l'hopital rule :

Problem 1 :

lim x-->0 (sin x/x)

Solution :

= lim x-->0 (sin x/x)

By applying the given limit, we get

= (sin 0/0)

= 0/0

Indeterminant form.

Differentiating the numerator and denominator using L'hopital rule, we get

= lim x-->0 (cos x / 1)

By applying the limit now

= cos 0/1

= 1/1

= 1

So, the value of lim x-->0 (sin x/x) is 1.

Problem 2 :

lim x-->0 (1 - cos 2x)/5x

Solution :

= lim x-->0 (1 - cos 2x)/5x

By applying the limit, we get

= (1 - cos 2(0))/5(0)

= (1 - cos 0)/0

= (1 - 1) / 0

= 0/0

Indeterminant form.

Differentiating the numerator and denominator, we get

= lim x-->0 (0 + 2 sin 2x)/5(1)

= lim x-->0 [2 sin 2x/5]

Applying the limit, we get

= 2 sin 2(0) / 5

= 2 sin 0 / 5

= 0/5

= 0

So, the value of lim x-->0 (1 - cos 2x)/5x is 0.

Should we apply L'Hopital rule ?

Problem 3 :

lim x-->0 sin x/(2x² - 1)

Solution :

= lim x-->0 sin x/(2x² - 1)

By applying the limit, we get

= sin 0/(2(0)² - 1)

= 0/-1

= 0

By evaluating the limit, we get 0 as result. It is defined value not an indeterminant form. Then there is no need to use L'Hopital rule.

So, the value of lim x-->0 sin x/(2x² - 1) is 0.

Problem 4 :

lim x-->0 sin x/(x² - 4)

Solution :

= lim x-->0 sin x/(x² - 4)

By applying the limit, we get

= sin 0/(0² - 4)

= 0/-4

= 0

By evaluating the limit, we get 0 as result. It is defined value not an indeterminant form. Then there is no need to use L'Hopital rule.

So, the value of  lim x-->0 sin x/(x² - 4) is 0.

Problem 5 :

lim x-->2 [sin (x - 2)/(x² - 4)]

Solution :

= lim x--> 2 [sin (x - 2)/(x² - 4)]

By applying the limit, we get

= lim x--> 2 [sin (2 - 2)/(2² - 4)]

= 0/0

Indeterminant form. Thus L'Hopital rule applies.

Differentiating the numerator and denominator, we get

= lim x--> 2 [cos (x - 2)/(2x - 0)]

= lim x--> 2 [cos (x - 2)/2x]

By applying the limit now, we get

= [cos (2 - 2)/2(2)]

= cos 0 / 4

= 1/4

So, the value of lim x-->2 [sin (x - 2)/(x² - 4)] is 1/4.

Problem 6 :

lim x-->1 [ln x/(x² - 1)]

Solution :

= lim x-->1 [ln x/(x² - 1)]

By applying the limit, we get

= [ln 1/(1² - 1)]

= 0/0

Indeterminant form. So, L'Hopital rule applies.

= lim x-->1 [(1/x)/(2x - 0)]

= lim x-->1 (1/2x²)

By applying the limit, we get

= 1/2(1)²

= 1/2

So, the value of lim x-->1 [ln x/(x² - 1)] is 1/2.

Problem 7 :

The table gives selected values of a twice differentiable function f(x)

Find lim x--> 3 [f(2x - 1)/x² - 9]

Solution :

= lim x--> 3 [f(2x - 1)/x² - 9]

Applying the limit, we get

= f(2(3) - 1) /(3² - 9)

= f(5) / 0

From the table, the value of f(5) is 0

= 0/0

Indeterminant form. Finding derivative since L'hopital rule applies.

= lim x--> 3 [f(2x - 1)'/(2x - 0)]

= lim x--> 3 [2f'(2x - 1)/(2x - 0)]

Applying the limit, we get

= [2f'(2(3) - 1)/2(3)]

= [2f'(5)/6]

= 2(7) / 6

= 7/3

Problem 8 :

lim x-->0 [(e^x - cos x - 2x)/(x² - 2x)]

a)  -1/2   b)  0       c)  1/2       d)  1

Solution :

lim x-->0 [(e^x - cos x - 2x)/(x² - 2x)]

Applying the limit directly,

= [(e⁰ - cos 0 - 2(0))/(0² - 2(0))]

= (1 - 1 - 0)/0

= 0/0

Indeterminant form.

Differentiating the numerator and denominator, we get

= lim x-->0 [(e^x + sin x - 2)/(2x - 2)]

= [(e⁰ + sin 0 - 2)/(2(0) - 2)]

= (1 + 0 - 2)/(0 - 2)

= -1/(-2)

= 1/2

So, option c is correct.

Problem 9 :

lim x--> ∞ ln(e^3x+ x) / x

a)  0      b)  1      c)  3     d)  ∞

Solution :

= lim x--> ∞ ln(e^3x+ x) / x

= lim x--> ∞ ln(e^∞+ ∞) / ∞

= ∞/∞

Indeterminant form.

= lim x--> ∞ (1/(e^3x+ x)) (3e^3x+ 1) / 1

= lim x--> ∞ ((3e^3x+ 1)/(e^3x+ x))

Applying the limit, again we get

= ∞/∞

Indeterminant form. Differentiating again,

= lim x--> ∞ ((9e^3x + 0)/(3e^3x + 1))

= lim x--> ∞ 9e^3x/(3e^3x + 1)

Differentiating, we get

= lim x--> ∞ 27e^3x/(9e^3x + 0)

= lim x--> ∞ 27e^3x/9e^3x

= 3

So, option c is correct.

Problem 10 :

The function f and h are twice differentiable with h(2) = 4. The function h satisfies h(x) = (x² - 4) / (1 - (f(x))³). , x ≠ 2. It is known that lim x --> 2 h(x) can be evaluated using L'Hopital's rule. Use lim x--> 2 h(x) to find f(2) and f'(2) 

Solution :

h(x) = (x² - 4) / (1 - (f(x))³)

Since h(x) is twice differentiable and it is continuous 

1 - f³(x) = 0

- f³(x) =  -1

f³(2) =  1

f(2) =  1

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