# 1.6.1 AP Calculus Exam Limits with L'H

• MHB
• karush
In summary, the conversation discusses the use of L'Hopital's rule to solve the limit $\displaystyle\lim_{x \to 0}\dfrac{1-\cos^2(2x)}{(2x)^2}$. The first step is to use the rule to find the derivative of the numerator and denominator, resulting in the limit $\displaystyle\lim_{x \to 0}\frac{4\sin(2x)\cos(2x)}{8x}$. This still results in an indeterminate form, so L'Hopital's rule is used again. After finding the second derivative, the limit becomes $\displaystyle\lim_{x \to 0}\frac{2(\cos karush Gold Member MHB$\displaystyle\lim_{x \to 0}\dfrac{1-\cos^2(2x)}{(2x)^2}=$by quick observation it is seen that this will go to$\dfrac{0}{0)}$so L'H rule becomes the tool to use but first steps were illusive the calculator returned 1 for the Limit$1-\cos^2 (2x) = \sin^2(2x)$you should be familiar with this basic limit ...$\displaystyle \lim_{u \to 0} \dfrac{\sin(u)}{u} = 1$been i while,,, i saw another example of the problem but expanded into confusion that was slam dunk mahalo If you really want to do it the "hard way", L'Hopital's rule does work. The derivative of$1- cos^2(2x)$is$4 sin(2x)cos(2x)$and the derivative of$(2x)^2= 4x^2$is$8x$so we look at the limit as x goes to 0 of$\frac{4 sin(2x)cos(2x)}{8x}= \frac{1}{2}\frac{sin(2x)cos(2x)}{x}$. The numerator and denominator of that still go to 0 so use L'Hopital's rule again! The derivative of$sin(2x)cos(2x)$is$2cos^2(2x)- 2sin^2(2x)$and the derivative of x is 1 so now we look at$\frac{1}{2}\frac{2(cos^2(x)- sin^2(x)}{1}$. At x= 0, cos(x)= 1 and sin(x)= 0 so that is$\frac{1}{2}(2(1- 0))= 1\$.
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## 1. What is the 1.6.1 AP Calculus Exam?

The 1.6.1 AP Calculus Exam is a standardized test administered by the College Board to assess a student's understanding of calculus concepts. It is typically taken by high school students who have completed a course in AP Calculus.

## 2. What is L'Hopital's rule?

L'Hopital's rule is a mathematical rule used to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞. It states that the limit of a quotient of two functions is equal to the limit of the quotient of their derivatives, under certain conditions.

## 3. How is L'Hopital's rule used in the 1.6.1 AP Calculus Exam?

L'Hopital's rule is frequently used in the 1.6.1 AP Calculus Exam to evaluate limits that would otherwise be difficult or impossible to solve. It allows students to find limits of indeterminate forms by taking the derivatives of the numerator and denominator and simplifying the resulting expression.

## 4. What are some common mistakes students make when using L'Hopital's rule?

One common mistake students make when using L'Hopital's rule is applying it incorrectly to limits that are not indeterminate forms. Another mistake is not checking the conditions for using L'Hopital's rule, such as making sure the limit is approaching 0 or ∞. Students may also make errors in taking the derivatives or simplifying the resulting expression.

## 5. How can I prepare for L'Hopital's rule questions on the 1.6.1 AP Calculus Exam?

To prepare for L'Hopital's rule questions on the 1.6.1 AP Calculus Exam, it is important to practice solving a variety of limits using this rule. Make sure to review the conditions for using L'Hopital's rule and double check your work for any mistakes. You can also seek help from a teacher or tutor if you are struggling with this concept.

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