2 votes 2 votes Which of the following limit is/are correct? $\displaystyle{}\lim _{x \rightarrow \infty} \sqrt[x]{x}=1$ $\displaystyle{}\lim _{x \rightarrow \infty} \sqrt[x]{x}=e$ $\displaystyle{}\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{x}=e^{2}$ $\displaystyle{}\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{x}=e$ Calculus goclasses2024-calculus-1 goclasses calculus limits multiple-selects 2-marks + – GO Classes asked Aug 28, 2022 retagged Apr 29, 2023 by Lakshman Bhaiya GO Classes 454 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
5 votes 5 votes $\begin{aligned} \lim _{x \rightarrow \infty} \sqrt[x]{x} &=\lim _{x \rightarrow \infty} x^{\frac{1}{x}} \\ &=\lim _{x \rightarrow \infty} e^{\frac{\ln x}{x}} \\ &=e^{\lim _{x \rightarrow \infty} \frac{\ln x}{x}} \\ &=e^{\lim _{x \rightarrow \infty} \frac{x^{-1}}{1}} \\ &=e^{\frac{0}{1}} \\ &=\boxed{1} \end{aligned}$ $\begin{aligned} \lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{x} &=\lim _{x \rightarrow \infty} e^{x \ln \left(1+\frac{2}{x}\right)} \\ &=e^{\lim _{x \rightarrow \infty} x \ln \left(1+\frac{2}{x}\right)} \\ &=e^{\lim _{x \rightarrow \infty} \frac{\ln \left(1+\frac{2}{x}\right)}{x^{-1}}} \\ &=e^{\lim _{x \rightarrow \infty} \frac{\left(1+\frac{2}{x}\right)^{-1}(-2) x^{-2}}{(-1) x^{-2}}} \\ &=e^{\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{-1}(2)} \\ &=e^{\left(1^{-1} \cdot 2\right)} \\ &=\boxed{e^{2}} \end{aligned}$ GO Classes answered Aug 28, 2022 edited Aug 29, 2022 by Lakshman Bhaiya GO Classes comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes I have problem in first part : Let me explain that. y = lim(x -> ♾️) (x^(1/x)) log(y) = lim(x -> ♾️)(log(x)/x) log(y) = lim(x -> ♾️)(1/x) : BY LH rule. log(y) = 0, now this is true when y = 1 only. I think it is true when y = e (stupid me..) Udhay_Brahmi answered Aug 31, 2022 Udhay_Brahmi comment Share Follow See all 0 reply Please log in or register to add a comment.