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 :

Solution :

Problem 2 :

Solution :

Factoring x2, we get

=  lim x-> ∞ x2(2 - 3/x2) / x2(1 - 5/x + 3/x2)

=  lim x-> ∞ (2 - 3/x2) / (1 - 5/x + 3/x2)

By applying the limit, we get

=  2/1

=  2

So, the answer is 2.

Problem 3 :

Solution :

By applying the limit, we get

=  ∞/log ∞

=  ∞/0

=  ∞

So, the answer is ∞.

Problem 4 :

Solution :

Using L'Hopital's rule, we get

So, the answer is 1.

Problem 5 :

Solution :

So, the answer is 0.

Problem 6 :

Solution :

By applying the limit, we get

=  0/2(1)

=  0

So, the answer is 0.

Problem 7 :

Solution :

So, the answer is -1.

Problem 8 :

Solution :

Let f(x)  =  xx

y  =  xx

log y  =  log xx

log y  =  x log x

log y  =  log x / (1/x)

lim x->0+ log y  =  lim x->0+ log x / (1/x)

lim x->0+ log y  =   ∞/(Indeterminant form)

Using L'Hopital's rule :

lim x->0+ log y  =  lim x->0+ (1/x) / (-1/x2)

lim x->0+ log y  =  lim x->0+ -x

By applying limit, we get

lim x->0+ log y  =  0

log (lim x->0+y)  =  0

(lim x->0+y)  =  e0

lim x->0+ y  =  1

So, the answer is 1.

Problem 9 :

Solution :

Let f(x)  =  (1 + (1/x))x

y  =  (1 + (1/x))x

log y  =  log (1 + (1/x))x

log y  =  x log (1 + (1/x))

log y  =  log (1 + (1/x)) / (1/x)

lim x -> log y  =  lim x ->∞ log (1 + (1/x)) / (1/x) 

By applying the limit, we get

lim x -> log y  =  lim x ->∞ log (1 + (1/)) / (1/)

lim x -> log y  =  lim x ->∞ log (1)/0

lim x -> log y  =  0/0 (Indeterminant form) 

By applying the limit, we get

lim x -> log y  =  1

log (lim x ->y)  =  1

lim x ->∞ y  =  e

Problem 10 :

Solution :

lim x->π/2  log y  =  lim x->π/2 (- sinx cosx)

By applying the limit, we get

lim x->π/2  log y  =  -sin(π/2) cos(π/2)

lim x->π/2  log y  =  1(0)

lim x->π/2  log y  =  0

log y (lim x->π/2  y)  =  0

(lim x->π/2  y)  =  1

So, the answer is 1.

Problem 11 :

Solution :

By applying the limit, we get

lim x->0+ log y  =  -1/2

log(lim x->0y)  =  -1/2

lim x->0y  =  e-1/2

lim x->0y  =  1/e

So, the answer is 1/e.

Problem 12 :

If an initial amount A0 of money is invested at an interest rate r compounded n times a year, the value of the investment after t years is

If the interest is compounded continuously, (that is as n->), show that the amount after t years is A = A0 ert

Solution :

A  =  A0 (1+ r/n)nt

By applying the limit, we get

log (lim n-> y)  =  rt

lim n-> y  =  ert

A = A0 ert

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