1
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

0
$\begingroup$

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:

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .