Use L'Hopital's rule to calculate derivatives.

[rizzi143]

Homework Statement

Use L'Hopital's rule to calculate the following derivatives.

Homework Equations

1. lim x-> pie/2 tan3x/tan5x

2. lim x->0 e^x - 1/sin x

3. lim x->1 e^x - e/In x

The Attempt at a Solution

i have attempted to solve the equations but differentiating the limits. but it seems i am not getting the right answer. the assignment is attached in the attachment. and it is question number 2 .

 

[Pengwuino]

Derivatives? These are just limits. Do you mean you need to use derivatives to calculate the limits?

 

[rizzi143]

Derivatives? These are just limits. Do you mean you need to use derivatives to calculate the limits?

well i am bit confused with this L'Hopital's rule. as far i understand the we need to get the derivative of the limits.

 

[Pengwuino]

L'Hopital's rule is used to find limits when of the indeterminate forms such as {{0} \over {0} exist when you initially take a limit. The assignment you have is poorly written. Those are not derivatives you are calculating, they are limits.

 

[rizzi143]

L'Hopital's rule is used to find limits when of the indeterminate forms such as {{0} \over {0} exist when you initially take a limit. The assignment you have is poorly written. Those are not derivatives you are calculating, they are limits.

right. that makes sense. to me .. did you check out the attachment? and question number 2 in that?

 

[Pengwuino]

right. that makes sense. to me .. did you check out the attachment? and question number 2 in that?

Yes, they are limits. I don't know why it says to to evaluate the derivatives. To determine those limits, you have to USE derivatives when using L'Hopital's rule, but in the end you're finding limits

 

[jhae2.718]

L'Hopital's rule states that, if \frac{f(x)}{g(x)} is an indeterminate form (i.e. 0/0, inf/inf) then \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}

So, you can take the derivative of the numerator and denominator, and if it is no longer an indeterminate form you can evaluate the limit. Otherwise, repeat until the limit is no longer an indeterminate form.

 

[rizzi143]

L'Hopital's rule states that, if \frac{f(x)}{g(x)} is an indeterminate form (i.e. 0/0, inf/inf) then \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}

So, you can take the derivative of the numerator and denominator, and if it is no longer an indeterminate form you can evaluate the limit. Otherwise, repeat until the limit is no longer an indeterminate form.

right. could you do first question for me to clarify.

 

[jhae2.718]

I won't do one of your questions for you (they're for you to practice after all!), but here's an example of how to apply L'Hôpital's Rule:

Consider the limit:
<br /> \lim_{x \to 0} x\ln{x}<br />
This is an indeterminate form of type 0 \cdot \infty, which means we can apply L'Hôpital's Rule.

First, we need to put it into the correct form of the fraction of two functions:
<br /> \lim_{x \to 0} x\ln{x} = \lim_{x \to 0} \frac{\ln{x}}{\frac{1}{x}}<br />
Take the derivatives of the numerator and denominator:
<br /> \lim_{x \to 0} \frac{\ln{x}}{\frac{1}{x}} = \lim_{x \to 0} \frac{(\ln{x})&#039;}{(\frac{1}{x})&#039;} = \lim_{x \to 0} \frac{\frac{1}{x}}{\frac{-1}{x^2}}<br />
This is no longer an indeterminate form, and simplifies to:
<br /> \lim_{x \to 0} -x<br />
This limit, of course, is 0.

 

[SammyS]

Use L'Hopital's rule to calculate the following derivatives.

Homework Equations

1. lim x-> pi/2 tan3x/tan5x

2. lim x->0 e^x - 1/sin x

3. lim x->1 e^x - e/In x

No doubt that's a typo & it should say: Use L'Hopital's rule to calculate the following limits.

I'll set-up the first:

\lim_{x\to\frac{\pi}{2}}\frac{\tan(3x)}{\tan(5x)}=\lim_{x\to\frac{\pi}{2}}\frac{\frac{d}{dx}\tan(3x)}{\frac{d}{dx}\tan(5x)}

That's just a start. It will take some trig. and two more applications of L'Hopital.