Vali said:Why the following limit doesn't exists ?
$$\lim_{x\rightarrow 0}xe^{-\frac{1}{x}}$$
I think it's because of $\frac{1}{x}$ which doesn't exists, right ?
No. As skeeter said, it is because the limits "from below" and "from above" are not the same. If the problem were $$\lim_{x\rightarrow 0^+}xe^{-\frac{1}{x}}$$ it would still be true that "the limit, as x goes to 0, of $\frac{1}{x}$ does not exist" but as $-\frac{1}{x}$ goes to negative infinit, $e^{-\frac{1}{x}}$ goes to 0. $$\lim_{x\rightarrow 0^+}xe^{-\frac{1}{x}}= 0$$.Vali said:Why the following limit doesn't exists ?
$$\lim_{x\rightarrow 0}xe^{-\frac{1}{x}}$$
I think it's because of $\frac{1}{x}$ which doesn't exists, right ?
Limits are used to determine the behavior of a function as the input approaches a certain value. However, if the function approaches different values from the left and right sides of the input, the limit does not exist.
A limit does not exist if the function approaches different values from the left and right sides of the input, or if the function approaches infinity or negative infinity.
Yes, a limit can fail to exist at a specific point if the function has a discontinuity or a vertical asymptote at that point.
When a limit does not exist, it means that the function does not have a defined value at that point. This could be due to various reasons, such as the function being undefined or approaching different values from different directions.
Yes, it is possible for a limit to not exist even if the function is continuous. This can happen if the function approaches different values from the left and right sides of the input, or if the function has a vertical asymptote at that point.