Answer:
The given question is asking for the limit of the function (x→0+) In(x²+1) / sin x². Let's break down the problem step by step to find the solution.
To find the limit as x approaches 0 from the positive side, we need to evaluate the function as x gets arbitrarily close to 0.
First, let's consider the numerator, ln(x²+1). The natural logarithm function, ln(x), is defined for positive values of x. Since x is approaching 0 from the positive side, the term x²+1 will always be positive. As x gets arbitrarily close to 0, x²+1 will approach 1. So, ln(x²+1) will approach ln(1), which is equal to 0.
Now, let's look at the denominator, sin x². As x approaches 0, x² will also approach 0. The sine function, sin(x), is defined for all real numbers. As x² approaches 0, sin x² will also approach 0.
Now, we can evaluate the limit by dividing the numerator and the denominator. We have 0 / 0, which is an indeterminate form. This means that we cannot directly evaluate the limit using basic algebra.
To solve this indeterminate form, we can apply L'Hôpital's Rule. This rule states that if we have an indeterminate form of the type 0 / 0 or ∞ / ∞, we can take the derivative of the numerator and the denominator separately and then evaluate the limit again.
Let's find the derivatives of the numerator and denominator:
The derivative of ln(x²+1) is (2x) / (x²+1).
The derivative of sin x² is cos x² * 2x.
Now, let's evaluate the limit of the derivatives:
lim x→0+ (2x) / (x²+1) / (cos x² * 2x)
As x approaches 0, the numerator 2x approaches 0 and the denominator (x²+1) approaches 1. The cos x² * 2x approaches cos(0) * 0, which is equal to 0.
So, we have 0 / 1 / 0, which is still an indeterminate form. We can apply L'Hôpital's Rule again.
Differentiating the numerator and denominator:
The derivative of 2x is 2.
The derivative of cos x² * 2x is -sin x² * (2x) + cos x² * 2.
Now, let's evaluate the limit of the derivatives again:
lim x→0+ 2 / (-sin x² * (2x) + cos x² * 2)
As x approaches 0, the numerator 2 remains constant. The denominator (-sin x² * (2x) + cos x² * 2) approaches (-sin(0) * (0) + cos(0) * 2), which is equal to 2.
So, we have 2 / 2, which simplifies to 1.
Therefore, the limit of the function (x→0+) ln(x²+1) / sin x² is 1.
Please let me know if there is anything else I can help you with.