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2 votes
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Which of the following limit is/are correct?

  1. $\displaystyle{}\lim _{x \rightarrow \infty} \sqrt[x]{x}=1$
  2. $\displaystyle{}\lim _{x \rightarrow \infty} \sqrt[x]{x}=e$
  3. $\displaystyle{}\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{x}=e^{2}$
  4. $\displaystyle{}\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{x}=e$
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2 Answers

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5 votes
$\begin{aligned} \lim _{x \rightarrow \infty} \sqrt[x]{x} &=\lim _{x \rightarrow \infty} x^{\frac{1}{x}} \\ &=\lim _{x \rightarrow \infty} e^{\frac{\ln x}{x}} \\ &=e^{\lim _{x \rightarrow \infty} \frac{\ln x}{x}} \\ &=e^{\lim _{x \rightarrow \infty} \frac{x^{-1}}{1}} \\ &=e^{\frac{0}{1}} \\ &=\boxed{1} \end{aligned}$

$\begin{aligned} \lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{x} &=\lim _{x \rightarrow \infty} e^{x \ln \left(1+\frac{2}{x}\right)} \\ &=e^{\lim _{x \rightarrow \infty} x \ln \left(1+\frac{2}{x}\right)} \\ &=e^{\lim _{x \rightarrow \infty} \frac{\ln \left(1+\frac{2}{x}\right)}{x^{-1}}} \\ &=e^{\lim _{x \rightarrow \infty} \frac{\left(1+\frac{2}{x}\right)^{-1}(-2) x^{-2}}{(-1) x^{-2}}} \\ &=e^{\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{-1}(2)} \\ &=e^{\left(1^{-1} \cdot 2\right)} \\ &=\boxed{e^{2}} \end{aligned}$
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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..)
Answer:

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