Verified answer

... ƒ(x)

= lim (x→∞) [ ( 3x³ + x² ) / ( bx³ + kx² ) ]^(x+1)

= lim (x→∞) [ x³ ( 3 + (1/x)) / x³( b + (k/x)) ]^(x+1)

= lim (x→∞) [ ( 3 + (1/x)) / ( b + (k/x)) ]^x • [ ( 3 + (1/x)) / ( b + (k/x)) ]^1

= lim (x→∞) { (3^x) [ 1 + (1/ 3x) ]^x / (b^x) [ 1 + (k/ bx) ]^x } • [ ( 3 + 0 ) / ( b + 0 ) ]

= lim (x→∞) [ (3/b)^x ] • [ ( e^(1/3)) / ( e^(k/b)) ] • ( 3/b ) ................ (1)

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Now, given that : b > 3.

∴ ( 3/b ) < 1

∴ | 3/b | = 3/b < 1

∴ lim (x→∞) (3/b)^x = 0 ................... (2)

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Using (2) in (1),

ƒ(x) = [ 0 ] • [ ... ] • [ 3/b ] = 0 .......................... Ans.

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Happy To Help !

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When you have a polynomial like this, the limit for x→∞ always goes toward the coefficient of the higher power, in this example 3/b, but to prove it, you factor:

([x³(3+1/x)]/[x³(b+k/x)])^(x+1) = ((3+1/x)/(b+k/x))^(x+1)

when x→∞, 1/x→0 and k/x→0, so you get:

f(x) = lim(x→∞) (3/b)^(x+1)

if b > 3, then 3/b < 1, then f(x) = 0

(If it were b < 3, then 3/b > 1, then f(x) = ∞)