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lim (2x^3+2)/(x^2+5) when x--> inf.
First we will divide by the highest power which is x^3.
==> lim(2x^3+2)/(x^2+5) = lim x^3(2+2/x^3)/lim x^3(1/x+5/x^3) = lim (2+2/x^3)/ lim (1/x +5/x^3) = (2+0)/(0+0)=2/0 = inf.
The limit `lim_(x->oo)(2x^3+2)/(x^2+5)` has to be determined.
If we substitute `x = oo` in `(2x^3+2)/(x^2+5)` , the numerator as well as the denominator are equal to `oo` which gives a result in the form oo/oo. This is indeterminate and allows the use of l'Hospital's rule to replace the numerator and denominator by their derivative.
The result is:
If we substitute `x = oo` in `(6x^2)/(2x)` the result is again `oo/oo` . The numerator and denominator have to be replaced by their derivatives again.
Now substituting `x = oo` gives the result `oo` .
The limit `lim_(x->oo)(2x^3+2)/(x^2+5) = oo` .
To find lt(2x^3+2)/(x^2+5), x->+infinity
Both numerator and denominator goes to infinity . But it is numerator which goes faster with higher exponent of the variable is one degree higher and hence limit must be infinity.
For calculation we devide numeartor and denominator by x^2 term by term and then set x--> infinity.
So lt (2x^3+2)/(x^2+5) = Lt (2x^3/x^2+2/x^2)/(2x^2/x^2+5/x^2) = lt (2x+2/x^2)(1+5/x^2) = (2*inf+0)(1+0) = infinity.
In order to calculate the limit of a rational function, when x tends to +inf., we'll divide both, numerator and denominator, by the highest power of x, which in this case is x^3.
lim (2x^3+2)/(x^2+5) = lim (2x^3+2)/lim (x^2+5)
lim (2x^3+2)/lim (x^2+5) = lim x^3*(2 + 2/x^3)/lim x^3*(1/x + 5/x^3)
After reducing similar terms, we'll get:
lim (2 + 2/x^3)/lim (1/x + 5/x^3) = (2+0)/(0+0)= 2/0= +inf.
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