- #1

agnimusayoti

- 240

- 23

- Homework Statement
- Find lim of this sequences as n approaching zero

- Relevant Equations
- $$\lim_{n\to\infty} (1+n^2)^\frac{1}{\ln n}$$

First I assume that

$$(1+n^2)^\frac{1}{\ln n}=\exp {\ln (1+n^2)^\frac{1}{\ln n}}

$$But,

$${\ln (1+n^2)^\frac{1}{\ln n}}={\frac{\ln (1+n^2)}{\ln n}}$$

By L'Hopital Rule, I got

$$\lim_{n\to\infty} {\frac{\ln (1+n^2)}{\ln n}}=\frac{\lim_{n\to\infty} (\frac{2n}{1+n^2})}{\lim_{n\to\infty} 1/n}$$

After simplification, by multiplying numerator by $$\frac{1/n}{1/n}$$, I got 2. But, I attempted to multiplying numerator by 1/n^2 since the highest power is 2. And I came with the result undefined because 0/0.

Can anyone explain what is my mistake? Which answer is true? Or these two answer were wrong?

Thanks a bunch, pals!

$$(1+n^2)^\frac{1}{\ln n}=\exp {\ln (1+n^2)^\frac{1}{\ln n}}

$$But,

$${\ln (1+n^2)^\frac{1}{\ln n}}={\frac{\ln (1+n^2)}{\ln n}}$$

By L'Hopital Rule, I got

$$\lim_{n\to\infty} {\frac{\ln (1+n^2)}{\ln n}}=\frac{\lim_{n\to\infty} (\frac{2n}{1+n^2})}{\lim_{n\to\infty} 1/n}$$

After simplification, by multiplying numerator by $$\frac{1/n}{1/n}$$, I got 2. But, I attempted to multiplying numerator by 1/n^2 since the highest power is 2. And I came with the result undefined because 0/0.

Can anyone explain what is my mistake? Which answer is true? Or these two answer were wrong?

Thanks a bunch, pals!