Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→0+ (sinx ln2x)
To find the limit as x approaches 0 from the right of the expression sin(x)ln(2x), we can use l'Hôpital's Rule if applicable. L'Hôpital's Rule states that if we have an indeterminate form (such as 0/0 or ∞/∞), we can differentiate the numerator and denominator separately to get an equivalent expression that may be easier to evaluate.
Let's apply l'Hôpital's Rule to this problem:
lim x→0+ (sin(x)ln(2x))
Both the numerator (sin(x)) and denominator (ln(2x)) approach 0 as x approaches 0, so this is an indeterminate form of 0/0.
Differentiating the numerator and denominator separately, we have:
lim x→0+ (cos(x)ln(2x) + sin(x) * (1/(2x))(2))
Simplifying this expression further, we get:
lim x→0+ (cos(x)ln(2x) + sin(x)/x)
Now, we can evaluate the limit as x approaches 0 from the right of this new expression.
When x approaches 0, both cos(x) and sin(x) approach 1, and sin(x)/x approaches 1 as well.
Therefore, the limit of the expression as x approaches 0+ is:
lim x→0+ (cos(x)ln(2x) + sin(x)/x)
= cos(0)ln(2*0) + sin(0)/0
= 0 + 0/0
= 0
So, the limit of sin(x)ln(2x) as x approaches 0+ is equal to 0.