To compute $\underset{n}{lim}(1+n{)}^{\frac{1}{\mathrm{ln}n}}$ without L'Hospital

doodverft05
2022-06-14
Answered

To compute $\underset{n}{lim}(1+n{)}^{\frac{1}{\mathrm{ln}n}}$ without L'Hospital

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Aaron Everett

Answered 2022-06-15
Author has **18** answers

I always treat limits of this form by computing the limit of the logarithm, so we want to consider

$\underset{n\to \mathrm{\infty}}{lim}\frac{\mathrm{ln}(1+n)}{\mathrm{ln}n}$

If we consider instead of the sequence the function limit

$\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{ln}(1+x)}{\mathrm{ln}x}$

we know that if this one exists, then it's the same of the limit of the sequence (the converse is generally not true). Now we can perform the substitution $x=1/t$ and note that

$\mathrm{ln}(1+1/t)=\mathrm{ln}(t+1)-\mathrm{ln}t$

so the limit becomes

$\underset{t\to {0}^{+}}{lim}\frac{\mathrm{ln}t-\mathrm{ln}(1+t)}{\mathrm{ln}t}=\underset{t\to {0}^{+}}{lim}{\textstyle (}1-\frac{\mathrm{ln}(1+t)}{\mathrm{ln}t}{\textstyle )}=1$

because the fraction is $0/\mathrm{\infty}$

Thus we have proved that

$\underset{n\to \mathrm{\infty}}{lim}(1+n{)}^{1/\mathrm{ln}n}={e}^{1}=e$

$\underset{n\to \mathrm{\infty}}{lim}\frac{\mathrm{ln}(1+n)}{\mathrm{ln}n}$

If we consider instead of the sequence the function limit

$\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{ln}(1+x)}{\mathrm{ln}x}$

we know that if this one exists, then it's the same of the limit of the sequence (the converse is generally not true). Now we can perform the substitution $x=1/t$ and note that

$\mathrm{ln}(1+1/t)=\mathrm{ln}(t+1)-\mathrm{ln}t$

so the limit becomes

$\underset{t\to {0}^{+}}{lim}\frac{\mathrm{ln}t-\mathrm{ln}(1+t)}{\mathrm{ln}t}=\underset{t\to {0}^{+}}{lim}{\textstyle (}1-\frac{\mathrm{ln}(1+t)}{\mathrm{ln}t}{\textstyle )}=1$

because the fraction is $0/\mathrm{\infty}$

Thus we have proved that

$\underset{n\to \mathrm{\infty}}{lim}(1+n{)}^{1/\mathrm{ln}n}={e}^{1}=e$

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