# Evaluate the limit at infinity: limit x--infinity (9e^(x)-3x)/(6e^x-4x) show the steps

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To find the lt x-->infinity (9e^x-3x)/(6e^x-4x)

We try putting x=0. The lt of the expression becomes (9e^if-3inf)(6e^inf - 4inf), which could not be simplified inthis form as infinity could not be treated like ordinary numbers. So it is an indeteminate position.

Let us divude both numerator and denominator, by e^x.

Then the expression,(9e^(x)-3x)/(6e^x-4x) becomes,(9-3x/e^x)/(6-4x/e^x).............(1)

Now, 3x/e^x = 3x/(1+x+x^2/2+x^3/3!++....). If you devide by x both numerator and denominators by x this expression becomes, 3/(1/x+1+x/2!+x^2/3!+....). Now if we put x=0, we get lt x-->03x/e^x = 3/(0+1+ inf/2+(inf)^2/3!+higher order terms} = 0

Similarly term ltx-->0 (4x/e^x) also equals zero.

With the preceding paras the The limit of the expression at (1) is (9-0)/(6-0) = 3/2

Aliter:We can use L' Hospial's rule also which covers the condition of inf/inf form indetrmination to determine the limits if the limit exists.