lim x->0 (e^x^2 - 1)/(square root(1+sinx)- square root(cosx))
To evaluate the limit of the given expression as x approaches 0, we can start by simplifying the expression.
Let's simplify the numerator first:
e^x^2 - 1
To simplify this, we can use the Maclaurin series expansion for the exponential function, which is given by:
e^x = 1 + x + (x^2 / 2!) + (x^3 / 3!) + ...
Using this expansion, we can rewrite e^x^2 as:
e^x^2 = 1 + x^2 + (x^4 / 2!) + (x^6 / 3!) + ...
Now let's simplify the denominator:
√(1 + sinx) - √cosx
We can't simplify this further, but we can make a simplification using the trigonometric identity:
√(1 + sinx) = √(1 + sin^2(x))
Using the identity sin^2(x) = 1 - cos^2(x), we can rewrite the numerator as:
√(1 + sinx) = √(1 + (1 - cos^2(x)))
= √(2 - cos^2(x))
Now, let's rewrite the given expression using these simplifications:
lim x->0 ((1 + x^2 + (x^4 / 2!) + (x^6 / 3!) + ...) - 1) / (√(2 - cos^2(x)) - √cosx)
Simplifying the numerator further, we get:
lim x->0 (x^2 + (x^4 / 2!) + (x^6 / 3!) + ...) / (√(2 - cos^2(x)) - √cosx)
Now, we can see that as x approaches 0, all the terms in the numerator approach 0 except for the x^2 term. Similarly, the denominator will also approach 0 as x approaches 0.
Therefore, we have an indeterminate form of 0/0. To evaluate the limit further, we can use L'Hôpital's Rule, which states that if we have a limit of the form 0/0 or ∞/∞, we can differentiate the numerator and denominator separately and then take the limit again.
Taking the derivative of the numerator and denominator, we have:
lim x->0 (2x + (4x^3 / 2!) + (6x^5 / 3!) + ...) / (√(2 - cos^2(x)) - √cosx)
Now, we can plug in x = 0 into the derivative, which gives us:
(2 * 0 + (4 * 0^3 / 2!) + (6 * 0^5 / 3!) + ...) / (√(2 - cos^2(0)) - √cos(0))
Simplifying, we get:
lim x->0 (0) / (√(2 - 1) - 1)
= 0 / (1 - 1)
= 0
Therefore, the limit of the given expression as x approaches 0 is 0.