Calculate limit (x^2+2x+1)/(2x^2-2x-1), x->+infinity
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limit (x^2+2x+1)/(2x^2-2x-1) when x--> inf
We observe that he highest power for the numerator and...
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The limit `lim_(x->oo)(x^2+2x+1)/(2x^2-2x-1)` has to be determined.
Substituting `x = oo` in the given expression `(x^2+2x+1)/(2x^2-2x-1)` gives the result `oo/oo` which is indeterminate. If the result obtained while determining limits is of the form `oo/oo` or `0/0` it is possible to use l'Hospital's rule and substitute the numerator and denominator by their derivatives.
The derivative of (x^2+2x+1) is 2x + 2 and the derivative of (2x^2-2x-1) is 4x - 2.
The given limit can be written as `lim_(x->oo) (2x + 2)/(4x - 2)`
If we substitute `x = oo` , we again get the indeterminate form `oo/oo` . Continue the earlier step and again substitute the denominator and numerator with their derivative.
This gives 2/4 = 1/2
The required limit `lim_(x->oo)(x^2+2x+1)/(2x^2-2x-1) = 1/2`
To find the lt(x^2+2x+1)/(2x^2-2x-1) as x-->+infinty.
Solution:
We divide term by term both numerator and denominator by x^2 and then take the limit. So.
{ lt(x^2+2x+1)/(2x^2-2x-1) as x-->+infinty} =
lt {1+2x/x^2+1/x^2)/(2-2x/x^2-1/x^2)} as x--> infinity.
Lt(1+2/x-1/x^2)/(2-2/x-1/x^2) as x--> infinity
= (1-2*0+0)/(2-2*0-0), 1/x and 1/x^2 approach zero when x-->inf.
=1/2
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^2.
We'll have:
lim[(x^2+2x+1)/(2x^2-2x-1)]= lim(x^2+2x+1)/lim(2x^2-2x-1)
lim x^2(1 + 2/x + 1/x^2)/lim x^2(2 - 2/x - 1/x^2)
After simplifying the similar terms, we'll get:
(lim 1 + lim 2/x + lim 1/x^2)/(lim 2 - lim 2/x - lim 1/x^2)
lim[(x^2+2x+1)/(2x^2-2x-1)]=(1+0+0)/(2-0-0)=1/2
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