For which pair of functions f(x) and g(x) below will the lim(x->infinity) f(x)g(x)≠0
a)f(x) = 10x + e^-x; g(x) = (1/5x)
b)f(x) = x^2; g(x) = e^-4x
c)f(x) = (Lnx)^3; g(x) = 1/x
d)f(x) = √x; g(x) = e^-x
exponentials grow and decay faster than polynomials. So, b,d -> 0
logs grow more slowly than polynomials, so c -> 0
a does not go to zero, since
f(x)g(x) = 2x^2 + x/5 e^-x -> 2x^2
To determine which pair of functions satisfies the condition lim(x->infinity) f(x)g(x)≠0, we need to analyze the behavior of the functions as x approaches infinity.
a) f(x) = 10x + e^-x; g(x) = (1/5x)
In this case, as x approaches infinity, both f(x) and g(x) tend to zero. Therefore, their product f(x)g(x) also approaches zero. So, (a) does not satisfy the condition.
b) f(x) = x^2; g(x) = e^-4x
For this pair, as x approaches infinity, g(x) = e^-4x approaches zero since the exponential function decays faster than any polynomial. However, f(x) = x^2 grows without bound as x approaches infinity. Therefore, f(x)g(x) approaches infinity, not zero. Thus, (b) satisfies the condition.
c) f(x) = (Lnx)^3; g(x) = 1/x
As x approaches infinity, g(x) = 1/x approaches zero. However, f(x) = (Lnx)^3 is a logarithmic function, which grows slowly and tends to infinity as x approaches infinity. Therefore, f(x)g(x) approaches infinity, not zero. So, (c) does not satisfy the condition.
d) f(x) = √x; g(x) = e^-x
In this case, as x approaches infinity, both f(x) = √x and g(x) = e^-x tend to zero. Hence, their product f(x)g(x) also approaches zero. Therefore, (d) does not satisfy the condition.
In conclusion, the pair of functions that satisfies the condition lim(x->infinity) f(x)g(x)≠0 is (b) f(x) = x^2; g(x) = e^-4x.
To determine which pair of functions will satisfy the condition lim(x->infinity) f(x)g(x) ≠ 0, we need to analyze the behavior of f(x) and g(x) as x approaches infinity.
a) f(x) = 10x + e^(-x); g(x) = (1/5x)
As x approaches infinity, both f(x) and g(x) tend towards infinity. Therefore, the product f(x)g(x) will approach infinity as well. So, lim(x->infinity) f(x)g(x) ≠ 0.
b) f(x) = x^2; g(x) = e^(-4x)
As x approaches infinity, g(x) = e^(-4x) approaches 0, while f(x) = x^2 approaches infinity. Therefore, the product f(x)g(x) will approach 0. So, lim(x->infinity) f(x)g(x) = 0.
c) f(x) = (ln(x))^3; g(x) = 1/x
As x approaches infinity, g(x) = 1/x approaches 0, while f(x) = (ln(x))^3 approaches infinity. Therefore, the product f(x)g(x) will approach 0. So, lim(x->infinity) f(x)g(x) = 0.
d) f(x) = √x; g(x) = e^(-x)
As x approaches infinity, both f(x) and g(x) tend towards 0. Therefore, the product f(x)g(x) will approach 0 as well. So, lim(x->infinity) f(x)g(x) = 0.
So, the only pair of functions for which lim(x->infinity) f(x)g(x) ≠ 0 is option a) f(x) = 10x + e^(-x); g(x) = (1/5x).