The discussion revolves around evaluating the limit \(\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}\). Participants explore various approaches to simplify and analyze the limit, focusing on the behavior of the functions involved as \(x\) approaches 2.
The conversation includes multiple interpretations of the limit, with some participants affirming the correctness of intermediate steps while others question them. There is a mix of agreement and uncertainty regarding the final result, with guidance offered on using L'Hôpital's rule and substitution methods.
Some participants highlight the importance of understanding the foundational concepts of limits and derivatives, suggesting that further review may be beneficial for those struggling with the material.
In English these are called indeterminate forms.Ssnow said:All it is correct but not necessary, I suggest you to put ##x=0## in the original limit, there are indefinite forms as ##\frac{0}{0},\frac{\infty}{\infty}, 0\cdot \infty, \infty-\infty## or not ?
The limit on the first line above can be evaluated merely by substituting x = 0. You don't need to any of the work you show on the following lines.gede said:How to solve this limit?
\lim_{x \to 0} \frac{\sqrt{x} (x - 7)}{\sqrt{x} - \sqrt{7}}
This is what I get:
\lim_{x \to 0} \frac{\sqrt{x} (x - 7)}{\sqrt{x} - \sqrt{7}} \frac{\sqrt{x} + \sqrt{7}}{\sqrt{x} + \sqrt{7}}
= \lim_{x \to 0} \frac{\sqrt{x} (x - 7) (\sqrt{x} + \sqrt{7})}{(x - 7)}
= \lim_{x \to 0} \sqrt{x}(\sqrt{x} + \sqrt{7})
What is the next solution?
Mark44 said:In English these are called indeterminate forms.