Question #31e17

2 Answers
Jim G.
Jun 29, 2017

#1#

Explanation:

#"divide terms on numerator/denominator by " x^2#

#rArrf(x)=((4x^2)/x^2-6/x^2)/((4x^2)/x^2-1/x^2)=(4-6/x^2)/(4-1/x^2)#

as #xto+oo,f(x)to(4-0)/(4-0)to1#

#rArrlim_(xto+oo)(4x^2-6)/(4x^2-1)=1#

Douglas K.
Jun 29, 2017

#lim_(xtooo) (4x^2-6)/(4x^2-1) = 1#

Explanation:

Given: #lim_(xtooo) (4x^2-6)/(4x^2-1) = ?#

Method 1:

Break into two fractions:

#lim_(xtooo) (4x^2-1)/(4x^2-1) -5/(4x^2-1)#

The first fraction becomes 1:

#lim_(xtooo) 1 -5/(4x^2-1)#

We know that the second term goes to zero as x goes to infinity, therefore, the limit is 1:

#lim_(xtooo) (4x^2-6)/(4x^2-1) = 1#

Method 2:

Because the expression evaluated at the limit is the indeterminate form #oo/oo# we can apply L'Hôpital's rule :

#lim_(xtooo) ((d(4x^2-6))/dx)/((d(4x^2-1))/dx) =#

#lim_(xtooo) (8x)/(8x) =#

#lim_(xtooo) 1 = 1#

Therefore, the limit of the original expression is 1:

#lim_(xtooo) (4x^2-6)/(4x^2-1) = 1#

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