Why can $\displaystyle \lim_{n \to \infty}\sqrt[n]{1+(\frac{1}{3})^n}=1$ be evaluated by first calculating the inner expression, while $\displaystyle\lim_{x \to +\infty}\frac{[(1+\frac{1}{x})^x]^x}{e^x}= \displaystyle\lim_{x \to +\infty}\frac{e^x}{e^x}=1$ is actually incorrect?
Asked
Active
Viewed 56 times
1
-
5Because it's an indeterminate form of type $\infty/\infty$ https://en.wikipedia.org/wiki/Indeterminate_form – Zima Jul 10 '24 at 09:14
1 Answers
0
As noticed in the comments, the key point is that the second one is an indeterminate form ($\infty/\infty$) while the first one is not ($1^0$).
Another example is the following
$$\lim_{x\to 0} \;(\cos x)^{\frac1{x^2}}$$
which is in the indeterminated form $1^{\infty}$ and solved wrongly gives the result (the exact result is indeed $e^{-\frac12}$)
$$\lim_{x\to 0} \;(\cos x)^{\frac1{x^2}}=\lim_{x\to 0} \;(1)^{\frac1{x^2}}=1$$
but for example the following
$$\lim_{x\to 0} \;(\sin x)^{\frac1{x^2}}$$
which is in the form $0^{\infty}$ (not indeterminated) can be solved in this way
$$\lim_{x\to 0} \;(\sin x)^{\frac1{x^2}}=\lim_{x\to 0} \;(0)^{\frac1{x^2}}=0$$
Refer also to:
user
- 162,563