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1. ### (Tricky) Absolute Value Inequalities

Thank you for your response, Mark44. Could you please explain the red box?
2. ### (Tricky) Absolute Value Inequalities

Hello everyone, I'm posting here since I'm only having trouble with an intermediate step in proving that \sqrt{x} \text{ is uniformly continuous on } [0, \infty] . By definition, |x - x_0| < ε^2 \Longleftrightarrow -ε^2 < x - x_0 < ε^2 \Longleftrightarrow -ε^2 + x_0 < x < ε^2 + x_0 1...
3. ### Limit of rational function to rational power

Thank you for your responses, Bohrok and HallsofIvy. @HallsofIvy: Thanks for your clarification on conjugates. How would you define a real conjugate then? Are two terms x and y conjugates of each other if and only x \times y are of degree 1 and do not have any fractional exponents?
4. ### Limit of rational function to rational power

Homework Statement Evaluate the limit, WITHOUT using l'Hôpital's rule: \lim_{x \rightarrow -1} \frac{x^{1/3} + 1}{x^{1/5} + 1} Homework Equations The Attempt at a Solution I tried to use the conjugate method which does not produce a useful outcome: \underset{x\to...
5. ### Limit of a trigonometric function (Involved problem)

Thank you very much for your replies, Tedjn and Bohrok. I was able to evaluate this limit.
6. ### Limit of a trigonometric function (Involved problem)

Homework Statement [B]Evaluate \underset{x\to 0}{\mathop{\lim }}\,\frac{\sec \frac{x}{2}-1}{x\sin x} , WITHOUT using l'Hôpital's rule. Homework Equations The Attempt at a Solution Hello there, I tried to evaluate this limit using two different approaches, both of which still...
7. ### Limit of a trigonometric function (Advanced problem)

Adminstrator: This is a double post, so please feel free to delete this one. The relevant post is "Limit of a Trigonometric Function (Involved Problem)". Homework Statement Evaluate \underset{x\to 0}{\mathop{\lim }}\,\frac{\sec \frac{x}{2}-1}{x\sin x} Homework Equations The Attempt at...
8. ### Limits of trig functions

Hi Bohrok, I'm just evaluating this limit for fun. Could you please reveal the last substitution which you made to get \frac{\sin x}{x} ? Thanks.
9. ### Proof that a limit does not exist with delta-epsilon definition

Thanks for your response, Hurkyl. Proof that \lim_{x \rightarrow 0} \frac{1}{x} does not exist: \lim_{x \rightarrow 0^{-}} \frac{1}{x} = \frac{1}{0^{-}} = -\infty \lim_{x \rightarrow 0^{+}} \frac{1}{x} = \frac{1}{0^{+}} = \infty Since the right- and left-sided limits differ and...
10. ### Proof that a limit does not exist with delta-epsilon definition

Thank you for your response, Hurkyl. However, how would I show with an assumed delta value that the above limit does not exist?
11. ### Proof that a limit does not exist with delta-epsilon definition

Hello there, I would like to learn how I can use the formal definition of a limit to prove that a limit does not exist. Unfortunately, my textbook (by Salas) does not offer any worked examples involving the following type of limit so I am not sure what to do. I write below that delta = 1 would...
12. ### Direction angles - Proof

Thanks for your reply, Dick. I have: \vec{u} = |\vec{u}| \cos A1 \hat{i} + |\vec{u}| \cos B1 \hat{j} + |\vec{u}| \cos Y1 \hat{k} \vec{v} = |\vec{v}| \cos A2 \hat{i} + |\vec{v}| \cos B2 \hat{j} + |\vec{v}| \cos Y2 \hat{k} If I dot these two expressions on the RS, I do not see...
13. ### Direction angles - Proof

Could anyone please offer an explanation for how to prove the above? Thank you!
14. ### Direction angles - Proof

Thank you for your response. Would you mind elaborating on the proof? I thought that the direction cosines themselves were the unit vectors, so how would \cos A1 \cos A2 = 0 ? Shouldn't the dot product of these direction cosines = 0?
15. ### Direction angles - Proof

Hello everyone, Thank you in advance for your help! --- Homework Statement 10. A vector \vec{u} with direction angles A1, B1, and Y1, is perpendicular to a vector \vec{v} with direction angles A2, B2, and Y2. Prove that: \cos A1 \cos B2 + \cos B1 \cos B2 + \cos Y1 \cos Y2 = 0. ---...
16. ### Derivative of y = cos(a^3 + x^3)

Hello there: Here is my solution, which matches your textbook's. y = \cos(a^3 + x^3) Let: u = a^3 + x^3 The Chain Rule: h'(x) = f'(g(x)) \times g'(x) Breaking the original equation down: f(x) = \cos u , f'(x) = -\sin u g(x) = u = a^3 + x^3 , g'(x) = 3x^2 (From the Product Rule)...