limit as x approaches infinity; tan^(-1) (x^2 - x^4)
any help is appreciated =D
I am going to use an intuitive approach
let theta = tan^-1 (x^2 - x^4)
or tan(theta) = x^2 - x^4
now as x---> ∞ the right side of the above equation ---> - ∞
I know that tan 90º or tan(pi/2) is undefined, but as I approach 90º or pi/2
the tangent value becomes infinitely large positively,
and the tangent of -90º or -pi/2 becomes -infinitely large.
so
limit tan^(-1) (x^2 - x^4) as x ---> ∞
= -pi/2
check:
set your calculator to radians
enter a huge number, multiply it by -1, then enter
2nd Tan
I got -1.570796 or - pi/2
To find the limit as x approaches infinity of the function f(x) = tan^(-1)(x^2 - x^4), we can use the properties of limits and some trigonometric identities.
First, let's rewrite the function using the arctan identity: tan^(-1)(x^2 - x^4) = π/2 - tan^(-1)(x^4 - x^2).
Now, as x approaches infinity, the term x^4 becomes much larger than x^2. This means that x^4 - x^2 will dominate the expression, and we can ignore the smaller x^2 term.
Therefore, the function can be approximated as f(x) ≈ π/2 - tan^(-1)(x^4).
Now, let's consider the limit as x approaches infinity of f(x):
lim(x -> ∞) [f(x)]
≈ lim(x -> ∞) [π/2 - tan^(-1)(x^4)]
≈ π/2 - lim(x -> ∞) [tan^(-1)(x^4)].
Now, let's focus on the term tan^(-1)(x^4). As x gets larger and larger, the value of x^4 also increases. We know that the arctan function has a range of (-π/2, π/2), which means that the value of tan^(-1)(x^4) will always be between (-π/2, π/2).
Therefore, as x approaches infinity, the value of tan^(-1)(x^4) approaches π/2. We can write this as:
lim(x -> ∞) [tan^(-1)(x^4)] = π/2.
Substituting this result back into the previous expression, we get:
lim(x -> ∞) [f(x)] ≈ π/2 - π/2
= 0.
So, the limit as x approaches infinity of tan^(-1)(x^2 - x^4) is 0.
I hope this helps! Let me know if you have any further questions.