To find Lt (x^2+1)/(x+1)^2 as x--> infinity.
As both numerator and denominator becomes infinity as x-->infinity . So it leads us to infinity/infinity form of indefiniteness.
So let f(x) = (x^2+1)/(x+1)^2
Then f(x) = (x^2+1)/(x^2+2x+1)
We divide both numerator and denominator bt x^2:
f(x) = (1+1/x^2)/(1+2/x+1/x^2)
Now we take the lt as x--> infinity.
Lt f(x) = Lt (1+1/x^2)/(1+2/x+1/x^2
Lt f(x) = (1+0)/(1+0+0) , as 1/x = 0 and 1/x^2 = 0 as x--> infinity.
lt f(x) = Lt (x^2+1)/(x+1)^2 = 1.
First, we'll expand the square from the denominator, using the formula:
We'll put a = x and b = 1.
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.
lim (x^2 + 1)/(x+1)^2 = lim (x^2 + 1)/lim (x^2+2x+1)
lim x^2*(1 + 1/x)/lim x^2*(1 + 2/x + 1/x^2)
After reducing similar terms, we'll get:
lim (x^2 + 1)/(x+1)^2 = (1)/(1+0)
lim (x^2 + 1)/(x+1)^2 = 1
We need to find limit of (x^2 + 1)/(x + 1)^2 , x--> + infinite.
Here we cannot replace x with infinity and get the result.
So we simplify the expression
(x^2 + 1)/(x + 1)^2
=> (x^2 + 1)/(x^2 + 2x + 1)
divide the numerator and denominator by x^2
=> [(1+1/x^2)] / [ 1 + 2/x + 1/x^2]
Now if limit x--> + infinity, we know that limit (1/x)--> 0
[(1+1/x^2)] / [ 1 + 2/x + 1/x^2]
=> (1+0)/ (1 + 0 + 0)
Therefore the required limit is 1.