# Thread: limit of a sequence as n approaching infinity

1. ## limit of a sequence as n approaching infinity

H'Lopital Application

2. Yes, its a L'Hospital's Rule's Application since its an indeterminate power ..
Personally, I prefer substitute $x=\sqrt{n}$ to make the differentiation step easier ..

If $x=\sqrt{n}$, clearly $x\to\infty$ as $n\to\infty$

so $\lim_{n\to\infty} (\sqrt{n}-1)^{\frac{1}{\sqrt{n}}}=\lim_{x\to\infty} (x-1)^{\frac{1}{x}}$ ..

You can evaluate the last one by the L'Hospital's Rule method for the indeterminate powers..
Or you can substitute $t=\frac{1}{x}$ to get a well-known limit ..

3. Originally Posted by General
Yes, its a L'Hospital's Rule's Application since its an indeterminate power ..
Personally, I prefer substitute $x=\sqrt{n}$ to make the differentiation step easier ..

If $x=\sqrt{n}$, clearly $x\to\infty$ as $n\to\infty$

so $\lim_{n\to\infty} (\sqrt{n}-1)^{\frac{1}{\sqrt{n}}}=\lim_{x\to\infty} (x-1)^{\frac{1}{x}}$ ..

You can evaluate the last one by the L'Hospital's Rule method for the indeterminate powers..
Or you can substitute $t=\frac{1}{x}$ to get a well-known limit ..

Is it e^0 = 1?

4. Correct.