Identify Limit: Identify the limit that needs to be evaluated.We need to find the limit of the function xsin(8x)−x2cos(x6) as x approaches 0.
Check Form: Check if the function is in an indeterminate form when x approaches 0. Plugging in x=0, we get (sin(0)−02cos(06))/0, which is of the form 0/0, an indeterminate form.
Apply L'Hôpital's Rule: Apply L'Hôpital's Rule since we have an indeterminate form of 0/0. L'Hôpital's Rule states that if the limit of f(x)/g(x) as x approaches a value c is of the form 0/0 or ∞/∞, then the limit is the same as the limit of f′(x)/g′(x) as x approaches c, provided that the latter limit exists.
Differentiate Numerator and Denominator: Differentiate the numerator and the denominator separately.The derivative of the numerator sin(8x) with respect to x is 8cos(8x).The derivative of the numerator −x2cos(x6) with respect to x is −2xcos(x6)+(x2)(x26)sin(x6) which simplifies to −2xcos(x6)+6sin(x6).The derivative of the denominator x with respect to x is 1.
Apply Derivatives to Rule: Apply the derivatives to L'Hôpital's Rule.Now we need to evaluate the limit of 18cos(8x)−2xcos(x6)+6sin(x6) as x approaches 0.
Evaluate Simplified Expression: Evaluate the limit of the simplified expression as x approaches 0. As x approaches 0, cos(8x) approaches cos(0) which is 1, and sin(x6) approaches sin(∞) which oscillates and does not have a limit. However, since sin(x6) is multiplied by x in the term 01, this term approaches 0. Therefore, the limit of the expression is 03, which simplifies to 04.
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