# What is the limit of f(x)/x^4 if x goes to infinite and f(x)=(x-1)(x-3)(x-5)(x-7) ?

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It is given that f(x)=(x-1)(x-3)(x-5)(x-7)

We have to find the value of lim x-->inf. [f(x)/x^4]

=> lim x-->inf. [(x-1)(x-3)(x-5)(x-7)/x^4]

=> lim x-->inf. [((x-1)/x)((x-3)/x)((x-5)/x)((x-7)/x)]

=> lim x-->inf. [(1 - 1/x)(1 - 3/x)(1 - 5/x)(1 - 7/x)]

As x tends to infinity, (1/x) tends to 0.

This gives (1 - 0)(1 - 0)(1 - 0)(1 - 0) = 1

**The required value of lim x-->inf. [f(x)/x^4] = 1**

We'll force the x factor in each pair of brackets of f(x), such as:

f(x) = x^4*(1 - 1/x)*(1 - 3/x)*(1 - 5/x)*(1 - 7/x)

We'll re-write the limit:

lim f(x)/x^4 = lim x^4*(1 - 1/x)*(1 - 3/x)*(1 - 5/x)*(1 - 7/x)/x^4

We'll simplify and we'll get:

lim f(x)/x^4 = lim (1 - 1/x)*(1 - 3/x)*(1 - 5/x)*(1 - 7/x)

Since the limit of each fraction 1/x ; 3/x ; 5/x ; 7/x, approaches to zero, when x approaches to infinite, we'll get:

lim f(x)/x^4 = (1 - 0)*(1 - 0)*(1 - 0)*(1 - 0)

lim f(x)/x^4 = 1

**The requested limit of the function f(x)/x^4, if x approaches to infinite, is lim f(x)/x^4 = 1.**