lim x->0 x^sin(x)
Same as lim x->0 x^x -> 0^0 = 1
d/dx[lnsquare root of(x^3) cube root of (x+3) fifth root of (3x-2)]
To find the limit of the function f(x) = x^sin(x) as x approaches 0, we can apply a well-known limit property called L'Hôpital's rule. L'Hôpital's rule states that if we have an indeterminate form of type 0/0 or ∞/∞, we can differentiate the numerator and denominator until we obtain a determinate form.
Let's start by applying L'Hôpital's rule to the given limit:
lim(x->0) x^sin(x)
Since we have an indeterminate form, we can differentiate the numerator and denominator with respect to x. Differentiating x^sin(x) requires using both the product and chain rule. We can rewrite the function as e^(sin(x)ln(x)) to make it easier to differentiate.
Taking the derivative of the numerator:
d/dx (x^sin(x)) = d/dx (e^(sin(x)ln(x)))
= e^(sin(x)ln(x)) * (cos(x)ln(x) + (sin(x)/x))
Taking the derivative of the denominator:
d/dx (x) = 1
Now, we can rewrite the limit as:
lim(x->0) [(e^(sin(x)ln(x))) * (cos(x)ln(x) + (sin(x)/x))] / 1
Since we have a determinate form of 0/1, we can evaluate the limit by substituting x = 0 into the expression above.
lim(x->0) [(e^(sin(x)ln(x))) * (cos(x)ln(x) + (sin(x)/x))] / 1
= (e^(sin(0)ln(0)) * (cos(0)ln(0) + (sin(0)/0)) / 1
= (e^(0*-∞)) * (1 * (-∞) + 0) / 1
= 0 * (-∞) / 1
= 0
Therefore, the limit of x^sin(x) as x approaches 0 is 0.