# limit of x as it approaches infinity (x^2+sinx)/(x^2+1)?

HELP please provide a clear and detailed explanation

Relevance

Look at the numerator. It fluctuates between x^2 - 1 and x^2 + 1 because sinx oscillates between -1 and +1.

Observe that the limit of (x^2 + 1) / (x^2 + 1) = 1 for all values of x.

Now look at (x^2 - 1) / (x^2 + 1) = (x^2 + 1 - 2) / (x^2 + 1) because x^2 - 1 = x^2 + 1 - 2.

(x^2 + 1 - 2) / (x^2 + 1) = (x^2 + 1) / (x^2 + 1) - 2/(x^2 + 1) = 1 - (2/(x^2 + 1))

As x approaches infinity, 1 - (2/(x^2 + 1) = 1 - 0 = 1.

So, we now have:

f(x) = (x^2 - 1) / (x^2 + 1)

g(x) = (x^2 + 1) / (x^2 + 1)

h(x) = (x^2 + sin x) / (x^2 + 1)

Observe that f(x) < h(x) < g(x)

Observe that lim f(x) = 1 = lim g(x) as x approaches infinity.

Therefore, the lim h(x) = 1 as x approaches infinity by the Squeeze Theorem

• Anonymous
7 years ago

1

• Lim x->~ (x^2+sinx)/(x^2+1) (devide all by x^2)

Lim x->~ 1+(sinx/x^2)/(1+1/x^2) = 1+0/1+0 = 1

remeber if sinx/x^2 will become 1/x

as sina/a = a/a = 1

so sinx/x^2 = x/x * 1/x =1/x which value of 1/x is 0

• Apply L Hospital rule and get limit x tends to infinity (2x + cos x)/2x. Split the limit to get limit x tends to infinity 1 which is 1, plus limit x tends to infinity cos x/x which is equal to product of 0 and a finite number between -1 and 1, that is 0. Hence answer is 1