Find limit x->x/2 tan^ sqroot (2sin^x+3sinx+4-sin^x+6sinx+2)
To find the limit as x approaches infinity of the given expression, we will simplify and evaluate step by step.
Step 1: Rewrite the expression
lim x->∞ [x/2 tan(sqrt(2sin^x + 3sin(x) + 4 - sin^x + 6sin(x) + 2))]
Step 2: Simplify the inner part of the expression
Inside the tangent function, we can see that there are two terms sin^x and 2sin^x. We can combine them as follows:
2sin^x + sin^x = 3sin^x
Similarly, combining the terms with sin(x):
3sin(x) + 6sin(x) = 9sin(x)
Now, the expression becomes:
lim x->∞ [x/2 tan(sqrt(3sin^x + 9sin(x) + 4 + 2))]
Step 3: Simplify the square root term
The square root term can be simplified by observing that both 3sin^x and 9sin(x) approach zero as x approaches infinity. Thus, we can ignore these terms:
sqrt(3sin^x + 9sin(x) + 4 + 2) ≈ sqrt(4 + 2) = sqrt(6)
Now, the expression becomes:
lim x->∞ [x/2 tan(sqrt(6))]
Step 4: Calculate the limit
We can now evaluate the limit. The tangent of a constant value, such as sqrt(6), is also a constant. Therefore, the limit is given by the constant multiplied by x/2 as x approaches infinity:
lim x->∞ [x/2 tan(sqrt(6))] = ∞/2 tan(sqrt(6)) = ∞ * tan(sqrt(6)) = ∞
So, the limit as x approaches infinity of the given expression is infinity.
You have very strange notation. I assume you mean
limit as x -> π/2 of
tan^2 √(2sin^2(x) + 3sinx + 4 - sin^2(x) + 6sinx + 2)
Clearly there is something wrong here. Try clearing it up, ok?
If I got it right, we have
tan^2(√(sin^2(x) - 3sinx + 2))
= tan^2(√((sinx-1)(sinx-2))))
as x -> π/2, that is
tan^2(0) = 0
So, I guess not.