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:

(a+b)^2=a^2+2ab+b^2

We'll put a = x and b = 1.

(x+1)^2=x^2+2x+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.

We'll have:

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

This gives

[(1+1/x^2)] / [ 1 + 2/x + 1/x^2]

=> (1+0)/ (1 + 0 + 0)

=> 1

**Therefore the required limit is 1.**