Find $\lim _{ n\rightarrow \infty }{ { \left( { 3 }^{ n }+1 \right) }^{ \frac { 1 }{ n } } } $ using the squeeze theorem

I have come across ways to do this but none mention the squeeze (or sandwich) theorem. I know I need to find $2$ functions which squeeze the given function but can only think of using $(3^n)^{1/n}$ i.e. $3$ as the $\le $ function


As you pointed out in the question, $(3^n+1)^{\frac{1}{n}}\geq 3$, and on the other hand $$ (3^n+1)^{\frac{1}{n}}\leq (3^n+3^n)^{\frac{1}{n}}=3\cdot 2^{\frac{1}{n}}$$ Since $\lim_{n\to\infty}2^{\frac{1}{n}}=1$, it follows that $\lim_{n\to\infty}(3^n+1)^{\frac{1}{n}}=3$ by the squeeze theorem.

  • $\begingroup$ Thank you so much! It's 4.40 am here and this question has been troubling me for hours, you're a life saver! $\endgroup$ – P Collier Jul 29 '16 at 18:41
  • $\begingroup$ No problem, the trick of turning addition into multiplication is a good one to remember. $\endgroup$ – carmichael561 Jul 29 '16 at 18:42


In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{1+x}\le \log(1+x)\le x} \tag 1$$

for $x>-1$.

Note that we can write

$$\begin{align} (1+3^n)^{1/n}&=3\left(1+\frac1{3^n}\right)^{1/n}\\\\ &=3e^{\frac1n \log\left(1+\frac{1}{3^n}\right)}\tag 2 \end{align}$$

Applying $(1)$ to $(2)$ we find that

$$\begin{align} 3e^{\frac{1}{n(1+3^n)}}\le (1+3^n)^{1/n}\le 3e^{\frac{1}{n3^n}} \tag 3 \end{align}$$

Finally, applying the squeeze theorem to $(3)$ reveals the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}(1+3^n)^{1/n}=3}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.