Limit of a function - Indeterminate Form and LH rule

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The discussion focuses on evaluating limits involving indeterminate forms using L'Hôpital's rule. The first limit, as x approaches 0, simplifies to an indeterminate form and is resolved through L'Hôpital's rule, yielding a final limit of 2. The second limit, as x approaches infinity, is also identified as an indeterminate form, prompting the need for further manipulation to find the solution. There is some confusion regarding the application of L'Hôpital's rule and the correct signs in the derivatives. Overall, the participants are working through the complexities of limits and the application of L'Hôpital's rule to resolve indeterminate forms.
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Homework Statement




Lim x-> 0 Sin(x) + Cos(x) + e^-x - 2 / Sin(x) - x



The Attempt at a Solution



Sin(x) + Cos(x) + e^-x - 2 = 0 and Sin(x) - x = 0

This is indeterminate form?

I used l'hospital's rule 3 times to get

-Cos(x) + Sin(x) - e^-x / -Cos(x)

The top equals -2 and the bottom equals -1. The limit is 2. Am I understanding this right?

Homework Statement



Lim x->infinity (sqrt(x^2+1) - 1)^ln(x)

The Attempt at a Solution



I'm not sure what to do with ln(x). I believe this is the indeterminate form infinity - infinity.
Am I supposed to rewrite this someway?
 
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should be -sinx, not +sinx after doing it three times.
But yes, the answer is 2.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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