Finding the Limit of tln(t) as t Approaches 0

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The limit of the function t ln(t) as t approaches 0 is a key topic in calculus. The discussion confirms that the derivative of sin(3x) is 3cos(3x) due to the chain rule, not 3sin(3x). Participants suggest exploring the minimum of the function h(t) = t ln(t) to understand its behavior near zero. This limit is commonly approached using L'Hôpital's Rule or by analyzing the function's critical points.

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  • Understanding of limits in calculus
  • Familiarity with derivatives and the chain rule
  • Knowledge of L'Hôpital's Rule
  • Basic concepts of function behavior and critical points
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  • Explore the behavior of logarithmic functions near zero
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mattmns
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Is the derivative of sin(3x) just 3sin(3x) because of the chain rule? IE, let u=3x, then 3sin(u) => 3sin(3x)

Thanks.
:cool:

Ok, I am pretty sure that is true. How about, <br /> <br /> \lim_{t\rightarrow 0} tln(t)<br /> <br />

What is the common approach for this problem?
 
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mattmns said:
Is the derivative of sin(3x) just 3sin(3x) because of the chain rule? IE, let u=3x, then 3sin(u) => 3sin(3x)

Thanks.
:cool:

Ok, I am pretty sure that is true. How about, <br /> <br /> \lim_{t\rightarrow 0} tln(t)<br /> <br />

What is the common approach for this problem?
It's not true because the derivative of sine is not itself

For the limit, you might look for a minimum of the function. Does it have one? More than one? That should get you started.
 
i agree with OlderDan.
h(x)=f(g(x))
if you have that, then:
h&#039;(x)=f&#039;(g(x))*g&#039;(x)
so, sine is not the derivative of itself.
 
I think the derivative of sine is cosine,
so the derivative of sin(3x) would be 3cos(3x)
 
Bah, that is what I meant, sorry. Thanks, been a while since I did a derivative or a limit :cry:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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