$\lim\limits_{n\to\infty}\frac{a^n}{1+na}$

Question

Let $g_n(x) = \dfrac{x^n}{1+nx}$ on $[0,1]$

I need to show that for every $a$ in $[0,1]$, $g_n(a)$ tends to $0$ as $n$ tends to infinity.

It is pretty much clear for $a = 0$ and $1$.

But if I take $a$ to be in $(0,1)$ for example $a = \dfrac{1}{2}$.

I will have to show that $\dfrac{\left(\frac{1}{2}\right)^n}{1+n\frac{1}{2}}$ tends to $0$.

I can see that the numerator is tending to $0$ and denominator to $\infty$.

I even tried using L'Hospital Rule. It is becoming really lengthy.

Help, please

Answer 1

Since $1+na \ge na$, $$\frac{a^n}{1+na} \le \frac{a^n}{na} =\frac{a^{n-1}}{n} \le \frac{1}{n} \to 0 $$

Answer 2

For $a\in(0,1)$ we have $$0\le g_n(a)\le a^n\xrightarrow{n\to\infty}0$$ Apply Sandwich theorem.

Answer 3

As the answers above have explained some elegant approaches to the problem, I would like to throw some light upon how to solve the sum using L's hospital rule, thus eliminating your doubt arising from what you have tried and was becoming "really lengthy".....

$\lim _{n\rightarrow \infty }\left(\frac{x^n}{1+xn}\right)=\lim _{n\rightarrow \infty }\left(\frac{\frac{d\left(x^n\right)}{dn}}{\frac{d\left(1+xn\right)}{dn}}\right)=\lim _{n\rightarrow \infty }\frac{x^n\ln x}{x}=\lim _{n\rightarrow \infty }\ln x\left(x^{n-1}\right)$

Now, as x lies between 0 and 1, the limit equals zero...

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There was no actual need of using L's hospital rule. Direct substitution would have given the answer easily. If any further queries please do ask.....

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P.S.-

L's hospital rule was absolutely unnecessary.I doubt whether we can use it,\ as 0/$\infty$ is not even an indeterminate form for x in between 0,1. But as the fraction would have been indeterminate if |x|>1, I suppose we could use L's hospital in this case...

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Edit- Thanks to Martund, who not only rectified my silly error,but also has helped eliminate any doubts. Here I would like to cite the main important points, which answer why L's hospital rule is applicable form Wikipedia:

• the rule applies to situations in which the ratio of the derivatives has a finite or infinite limit, but not to situations in which that ratio fluctuates permanently as x gets closer and closer to c.- In our case, the ratio doesn't fluctuate.
• The requirement that the limit ${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$ exists is essential- in this case, both the limits exist.