I have a test soon.. and I really need to know how to do this problem.. please help!!!
lim as x-->0 sin^2(x)/tan(x^2)
the answer is 1, but I have no clue how to get that!
Use series expansions. Look up Taylor expansion on google and study that topic first.
Using series expansion techniques these limit problems become pretty trivial. For x in a neighborhood of zero:
Sin(x) = x + O(x^3) ---->
sin^2(x) = x^2 + O(x^4)
tan(x) = x + O(x^3) -->
tan(x^2) = x^2 + O(x^6)
And you see that:
sin^2(x)/tan(x^2) = 1 + O(x^4)
So, the limit is 1
To solve this problem, you can use series expansions to simplify the expression. A series expansion is a way to represent a function as an infinite sum of terms. In this case, we can use the Taylor expansion to approximate the functions involved.
First, let's look at the series expansion of sine function:
sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
Using only the first term of the expansion, we have:
sin(x) ≈ x
Next, let's look at the series expansion of tangent function:
tan(x) = x + (x^3)/3 + (2x^5)/15 + (17x^7)/315 + ...
Again, using only the first term of the expansion, we have:
tan(x) ≈ x
Now, let's apply these approximations to the expression:
sin^2(x)/tan(x^2)
≈ (x^2)/(x^2)
= 1
Therefore, the limit as x approaches 0 of sin^2(x)/tan(x^2) is equal to 1.
This approach of using series expansions is a powerful tool to simplify complicated mathematical expressions and evaluate limits. It's always useful to study and understand Taylor expansions to better tackle such problems.