Let $u = \frac{1}{\log (x)}$. If $u \to 0$. Then find the value of $x$.

Asked Mar 06 '22 at 09:47

Active Mar 06 '22 at 10:55

Viewed 35 times

My attempt: $u = \frac{1}{\log (x)}$. When $u\to0$ then $0 = \frac{1}{\log (x)} \implies (\log x)^{-1} = 0 \implies (0)^{-1} = \log (x)\implies \log (x) = 0 \implies x = e^0 \implies x = 1$ but answer is $x = 0$. Please help me. How to solve this type of problems.

edited Mar 06 '22 at 10:55

Vivaan Daga

5,531

asked Mar 06 '22 at 09:47

Poonam Bai

1

Hint: $1/f \to 0$ iff $|f|\to \infty$. – Evangelopoulos Foivos Mar 06 '22 at 09:49
the "value" of x? – Mr.Gandalf Sauron Mar 06 '22 at 09:50
Well,the answer must be no such x exists.. – Mar 06 '22 at 09:51
Hint : We have $$x=e^{\frac{1}{u}}$$ What happens if $u\rightarrow 0$ a) from the right and b) from the left ? – Peter Mar 06 '22 at 09:52
The inverse of $0$ is not $0$, we have $0\cdot 0=0\ne 1$. In fact, $0$ is not a unit , hence has no inverse at all. – Peter Mar 06 '22 at 09:56

Let $u = \frac{1}{\log (x)}$. If $u \to 0$. Then find the value of $x$.

0 Answers0