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simple calculus question

1. Evaluate the following limits.

(a) lim x^2/(cos9x-cos5x) =_______
x->0

(b) lim x^(1/3) ln x =_______
x->0+

(c) lim ((1/ln x) - (1/x-1)) =_________
x->1+

(d) lim ((1 + (5/x))^(x/2) =__________
x->+infiniti

Solution Preview

1.

lim_(x->0) x^2/(cos(9 x)-cos(5 x))

Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->0) x^2/(-cos(5 x)+cos(9 x)) = lim_(x->0) (( dx^2)/( dx))/(( d(-cos(5 x)+cos(9 x)))/( dx)):
= lim_(x->0) (2 x)/(5 sin(5 x)-9 sin(9 x))

Factor out constants:
= 2 (lim_(x->0) x/(5 sin(5 x)-9 sin(9 x)))

Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(x->0) x/(5 sin(5 x)-9 sin(9 x)) = lim_(x->0) (( dx)/( dx))/(( d(5 sin(5 x)-9 sin(9 x)))/( dx)):
= 2 (lim_(x->0) 1/(25 cos(5 x)-81 cos(9 x)))

The limit of a quotient is the quotient of the limits:
= 2/(lim_(x->0) (25 cos(5 x)-81 cos(9 x)))

The limit of a sum is the sum of the limits:
= 2/(25 (lim_(x->0) cos(5 x))-81 (lim_(x->0) cos(9 x)))

Using the continuity of cos(x) at x = 0 write lim_(x->0) cos(5 x) as cos(lim_(x->0) 5 x):
= 2/(25 cos(lim_(x->0) 5 x)-81 (lim_(x->0) cos(9 x)))

Using the continuity of cos(x) at x = 0 write lim_(x->0) cos(9 x) as cos(lim_(x->0) 9 x):
= 2/(25 cos(lim_(x->0) 5 x)-81 cos(lim_(x->0) 9 x))

Factor out constants:
= 2/(25 cos(5 (lim_(x->0) x))-81 cos(lim_(x->0) 9 x))

The limit of x as x approaches 0 is 0:
= 2/(25 cos(0)-81 cos(lim_(x->0) 9 x))

Factor out constants:
= 2/(25 cos(0)-81 cos(9 (lim_(x->0) x)))

The limit of x as x approaches 0 is 0:
= -1/28

2.

lim_(x->0) x^(1/3) log(x)

Indeterminate form of type 0·infinity. Let t = 1/x, then lim_(x->0) x^(1/3) log(x) = lim_(t->infinity) (1/t)^(1/3) log(1/t):
= lim_(t->infinity) (1/t)^(1/3) log(1/t)

Indeterminate form of type 0·infinity, write lim_(t->infinity) (1/t)^(1/3) ...

Solution Summary

simple calculus question

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