EVALUATE THE FOLLOWING LIMIT USING L'HOPITAL'S RULE

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)  =  x^x

y  =  x^x

log y  =  log x^x

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->0⁺ y)  =  -1/2

lim x->0⁺ y  =  e^-1/2

lim x->0⁺ y  =  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 e^rt

Solution :

A  =  A₀ (1+ r/n)^nt

By applying the limit, we get

log (lim n->∞ y)  =  rt

lim n->∞ y  =  e^rt

A = A₀ e^rt

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