Saved Bookmarks
| 1. |
(x) \( f(x)=\left\{\begin{array}{r}(x-a)^{2} \cos \frac{1}{x-a}, x \neq a \\ 1, x=a\end{array}\right. \) at \( x=a \) |
|
Answer» \( f(x) = \begin{cases} (x-a)^2 & cos\,\frac{1}{x-a},x≠a\\ & 1,x=a\\ \end{cases} \) Now, \(\lim_{x\to a}\) f(x) = \(\lim_{x\to a}\) (x - a)2 cos\((\frac{1}{x-a})\) = \(\lim_{x\to a}\) (x - a)2 x l (where l = \(\lim_{x\to a}\) cos\((\frac{1}{x-a})\) ∴ l \(\leq\) l \(\leq\) 1 (bounded value) = (a - a)2 x l = 0 x l = 0 \(\neq\) 1 \(\lim_{x\to a}\) f(x) \(\neq\) f(a) \(\Rightarrow\) f(x) is not continuous at x = a |
|