# Evaluating the Limit of Cosine Function Using L'Hospital's Rule - Explained

• navneet9431
In summary: A)}{x^2}=\frac{\sin(\pi/2 \: A)}{x^2}-\frac{1}{x^2}\frac{\cos(\pi/2 \: A)}{x^2}.$$So$$\frac{\cos(\pi/2 \: A)}{x^2}-\frac{1}{x^2}\frac{\cos(\pi/2 \: A)}{x^2}=0$$This is true for all ##A##. navneet9431 Gold Member <Moderator's note: Moved from a technical forum and thus no template.>$$\lim_{x \to 0} \cos(\pi/2\cos(x))/x^2$$I tried to evaluate the limit this way,$$\lim_{x \to 0} \cos(\pi/2\cdot1)/x^2$$since$$\cos0=1\lim_{x \to 0} \cos(\pi/2\cdot1)/x^2=\lim_{x \to 0} 0/x^2$$Now apply L'Hospital's Rule twice,$$\lim_{x \to 0} 0/2(x)=\lim_{x \to 0} 0/2=0$$So,this way the answer is zero. Can you please explain where am I doing wrong? I will be thankful for help! Last edited by a moderator: You cannot just take the limit for one part and leave the rest. Either you take the limit for the whole expression or not. This would be long, but use the L'Hopital rule twice on the original expression to remove x^2 from the denominator navneet9431 Phylosopher said: You cannot just take the limit for one part and leave the rest. Either you take the limit for the whole expression or not. This would be long, but use the L'Hopital rule twice on the original expression to remove x^2 from the denominator It s enough to apply L'Hopital rule once. You will get a product, and arrange the factors, so they have known limits. (One factor will be sin(x)/x.) SammyS, Delta2 and navneet9431 The product rule for limits ##\lim{f(x)g(x)}=\lim{f(x)}\lim{g(x)}## holds only if both the limits of ##f(x)## and ##g(x)## are finite or ##+-\infty##. If one limit is 0 and the other is infinite, as happens in this case where ##\lim_{x\to 0}\cos(\frac{\pi}{2}\cos x)=0## and ##\lim_{x\to 0}{\frac{1}{x^2}}=\infty##, then the product rule cannot be applied. Last edited: Phylosopher navneet9431 said: <Moderator's note: Moved from a technical forum and thus no template.>$$\lim_{x \to 0} \cos(\pi/2\cos(x))/x^2$$I tried to evaluate the limit this way,$$\lim_{x \to 0} \cos(\pi/2\cdot1)/x^2$$since$$\cos0=1\lim_{x \to 0} \cos(\pi/2\cdot1)/x^2=\lim_{x \to 0} 0/x^2$$Now apply L'Hospital's Rule twice,$$\lim_{x \to 0} 0/2(x)=\lim_{x \to 0} 0/2=0$$So,this way the answer is zero. Can you please explain where am I doing wrong? I will be thankful for help! If you write ##\cos x = 1- A## for small ##|x|##, your given fraction is$$\text{fraction} = \frac{\sin(\pi/2 \: A)}{x^2}. Now look at how ##A = 1 - \cos x## behaves for small ##|x|##, or use l'Hospital's rule.

## 1. What is L'Hospital's Rule?

L'Hospital's Rule is a mathematical theorem that allows for the evaluation of limits involving indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of the quotient of two functions f(x) and g(x) is indeterminate, then the limit of their derivatives f'(x) and g'(x) will be the same.

## 2. How is L'Hospital's Rule used to evaluate the limit of cosine function?

L'Hospital's Rule can be applied to the limit of cosine function by taking the derivative of both the numerator and denominator and then evaluating the limit again. This process can be repeated until the limit is no longer indeterminate, and the final result will be the same as the original limit.

## 3. Why is L'Hospital's Rule useful for evaluating limits?

L'Hospital's Rule is useful because it provides a method for evaluating limits that would otherwise be impossible or very difficult to solve. It is especially helpful for evaluating limits involving trigonometric functions, exponential functions, and logarithmic functions.

## 4. Are there any limitations to using L'Hospital's Rule?

Yes, there are limitations to using L'Hospital's Rule. It can only be used for limits involving indeterminate forms, and it may not always provide a definitive answer. In some cases, it may lead to an infinite loop of taking derivatives, and the limit may never be reached.

## 5. Can L'Hospital's Rule be used for all types of functions?

No, L'Hospital's Rule is not applicable to all types of functions. It can only be used for functions that are differentiable, meaning they have a well-defined derivative at the point where the limit is being evaluated. Additionally, it can only be used for limits that involve indeterminate forms.

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