Question

Find the limit. Use L'Hospital's rule where appropriate. If there is an applicable alternate method to L'Hospital's rule, consider using it instead. Find the limit as x approaches 0 of (7x² sin(x))/(sin(x) - x)?

Answer 1

To find the limit as x approaches 0 of (7x² sin(x))/(sin(x)-x), we can apply L'Hospital's rule. After applying L'Hospital's rule multiple times and simplifying, we find that the limit is 0.

Step-by-step explanation:

To find the limit as x approaches 0 of (7x² sin(x))/(sin(x)-x), we can apply L'Hospital's rule. First, let's evaluate the limit as x approaches 0 of the numerator, 7x² sin(x). Using the product rule, we can differentiate 7x² sin(x) with respect to x to get 14x sin(x) + 7x² cos(x). Similarly, let's evaluate the limit of the denominator, sin(x) - x, as x approaches 0. Applying the sine rule and derivative of x with respect to x, we get cos(x) - 1.

Now, we can rewrite the original limit as the limit as x approaches 0 of (14x sin(x) + 7x² cos(x))/(cos(x)-1). Applying L'Hospital's rule again, we differentiate the numerator and the denominator. Taking the derivative of 14x sin(x) + 7x² cos(x) with respect to x, we get 14sin(x) + 14x cos(x) + 14x cos(x) - 14x² sin(x). Derivative of cos(x) - 1 is -sin(x).

We can simplify the expression in the as x approaches 0 to (14sin(x) + 14x cos(x) + 14x cos(x) - 14x² sin(x))/(-sin(x)). Canceling out common terms and using the fact that sin(0) = 0 and cos(0) = 1, we are left with 28x/(-sin(x)). Taking the limit as x approaches 0, we substitute 0 for x to find that the limit is 0.