# Evaluate the limit of function (7x^2+5x)/(8x^3+6x), x-->infinity.

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### 3 Answers

To evaluate the limit first we will divide by the highest power of the function which is x^3.

==> lim (7x^2+5x)/(8x^3+6x) = lim x^3(7/x+5/x^2)/limx^3(8+6/x^2)

==> lim (7/x +5/x^2)/ lim (8+6/x^2) = 0+0/(8+0)=0/8=0

To find the lt(7x^2+5x)/(8x^3+6x), x-->infinity.

Solution:

As x approaches infinity both numerator and denominator goes to infinity. Being an indetrminate of the infinity/infinity form this could be solved by L' Hospial's rule of diffrentiating numerator and denominator and then taking the limit or dividing numerator and denominator by x^3 term by term and then taking the limit.

lt(7x^2+5x)/(8x^3+6x), x-->infinity.

= Lt (7x^2/x^3+5x/x^3)/(8x^3/x^3+6x/x^3) as x-->inf

=Lt(7/x+5/x^2)/(8+5/x^2) = (7*0+0)/(8+0) = 0/8 = 0

To evaluate the limit of the rational function, when x tends to +inf.,we'll factorize both, numerator and denominator, by the highest power of x, which in this case is x^3.

We'll have:

lim (7x^2+5x)/(8x^3+6x) = lim (7x^2+5x)/lim (8x^3+6x)

lim (7x^2+5x)/lim (8x^3+6x) = lim x^3*(7/x + 5/x^2)/lim x^3*(8 + 6/x^2)

After reducing similar terms, we'll get:

lim (7/x + 5/x^2)/lim (8 + 6/x^2)= (0+0)/(8+0)= 0/8= 0.