The limit problem discussed is \lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}, which evaluates to - \frac{1}{4}. The solution involves applying L'Hôpital's rule and recognizing the limit \lim_{t \to 0} \frac{\tan t}{t} = 1. Participants confirmed the correctness of the solution and discussed alternative methods, including substitution techniques and the importance of understanding derivatives for applying L'Hôpital's rule effectively.
\lim_{t \to 0} \frac{\tan t}{t} = 1Students and educators in calculus, mathematicians solving limit problems, and anyone looking to deepen their understanding of trigonometric limits and L'Hôpital's rule.
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.