# When a function contains a sequence, and how to find the function's limit?

Suppose $\lim \limits_{n \to \infty} a_n = 0$. Find the limit

$$\lim \limits_{n \to \infty} \left(1+a_n \frac{x}{n}\right)^n$$

It's kind intuitive that the answer is 1, but clearly I can't just say that the limits equal $\lim \limits_{n \to \infty} 1^n = 1$.

I feel like I should let $y =(1+a_n \frac{x}{n})^n$. Then take $log$ so that $log(y) = nlog(1+a_n \frac{x}{n})$ But L'hopital doesn't apply here either

• hint :$$n\log(1+a_n \frac{x}{n}) = \dfrac{\log(1+a_n \frac{x}{n})}{1/n}$$ – AgentS Mar 18 '15 at 14:25
• @ganeshie8 so we apply limit, and using L'hopital rule. But know do I take the derivative of a sequence? – Ellery Lai Mar 18 '15 at 14:27
• You definitely can't say it is like $\lim_{n\to\infty} 1^n$, because that same reasoning would show that $\lim (1+x/n)^n$ is $1$, when it is $e^x$. You need that $a_n\to 0$. – Thomas Andrews Mar 18 '15 at 14:29

If $a_n \to 0$ then, no matter how big $x$ is, for any $n$ big enough we have $|a_n x|\leq \varepsilon$, hence: $$\left(1-\frac{\varepsilon}{n}\right)e^{-\varepsilon}\leq\left(1+\frac{a_n x}{n}\right)^n \leq e^{\varepsilon}.$$ Since $\varepsilon$ is an arbitrary positive number, the claim follows by squeezing.

• Could you explain why is $$\lim \limits_{n \to \infty} e^{\epsilon} = 0?$$ I know that $\epsilon$ can be arbitrary small positive, but why does it have anything to do with $n \to \infty ?$ – Ellery Lai Mar 18 '15 at 14:50
• @ElleryLai: I have not stated that. The point is the following: given any positive $\varepsilon$, the sequence given by $b_n=\left(1+\frac{a_n x}{n}\right)^n$, for any $n$ big enough, is between the LHS and the RHS of what I wrote, hence: $$\lim_{n\to +\infty} b_n = \color{red}{1}.$$ – Jack D'Aurizio Mar 18 '15 at 15:00
• I couldn't figure out why is that $$\left(1+\frac{a_n x}{n}\right)^n \leq e^{\varepsilon}.$$ ?? because of the limit definition of limit? – Ellery Lai Mar 18 '15 at 15:30
• @ElleryLai: because for any positive $A$, the sequence $$(1+A/n)^n$$ is an increasing sequence that converges towards $e^A$. Basic fact. – Jack D'Aurizio Mar 18 '15 at 15:33
• Since $-\epsilon < a_{n}x < \epsilon$ the inequality should be like $$\left(1 - \frac{\epsilon}{n}\right)^{n} < \left(1 + \frac{a_{n}x}{n}\right)^{n} < \left(1 + \frac{\epsilon}{n}\right)^{n}$$ and this would lead to $$e^{-\epsilon} \leq \left(1 + \frac{a_{n}x}{n}\right)^{n}\leq e^{\epsilon}$$ for sufficiently large values of $n$. – Paramanand Singh Apr 4 '15 at 16:43

Hint: Supose $\lim_{n\to \infty}n\cdot\frac{1}{ x\cdot a_n}=0$. Note that $\lim_{n\to \infty} x\cdot a_n=0$, $$\left( 1+a_n\frac{x}{n} \right)^{n} = \left( 1+\frac{1}{n\cdot\frac{1}{ x\cdot a_n}} \right)^n = \left[ \left( 1+\frac{1}{n\cdot\frac{1}{ x\cdot a_n}} \right)^{n\cdot\frac{1}{\color{red}{ x\cdot a_n}}} \right]^{\color{red}{ x\cdot a_n}}$$ and $\lim_{n\to \infty} \left( 1+\frac{1}{\color{blue}{n\cdot\frac{1}{ x\cdot a_n}}} \right)^{\color{blue}{n\cdot\frac{1}{ x\cdot a_n}}}=e$. If $\lim_{n\to \infty}n\cdot\frac{1}{ x\cdot a_n}=0$ then $$\lim_{n\to \infty} \left[ \left( 1+\frac{1}{n\cdot\frac{1}{ x\cdot a_n}} \right)^{n\cdot\frac{1}{\color{red}{ x\cdot a_n}}} \right]^{\color{red}{ x\cdot a_n}} =\ldots =e^0=1$$ The case $\lim_{n\to \infty}n\cdot\frac{1}{ x\cdot a_n}=L\in\mathbb{R}$ is trivial.

• why is that $\lim \limits_{n \to \infty} \frac{n}{xa_n} = 0 ?$ – Ellery Lai Mar 18 '15 at 14:56
• also, for you last part, you seem to use if $$\lim \limits_{n \to \infty} f(n) = l$$ and $$\lim \limits_{n \to \infty} g(n) = m$$, then $$\lim \limits_{n \to \infty} f(n)^{g(n)} = l^m$$. I'm not sure if this is one of the limit properties – Ellery Lai Mar 18 '15 at 16:24
• @Ellery If $\lim_{n\to \infty}\frac{n}{x\cdot a_n}=L\neq 0$ and $L\in\mathbb{R}$ then the question is trivial. In your secont coment you are right. I hope helped you. – Elias Costa Mar 18 '15 at 16:28
• @ElleryLai Do not intend to exhaust all cases to be analyzed. In the case of home work, is not purpose of this site provide full answers. – Elias Costa Mar 18 '15 at 16:37