JEE Limits 1

First off all, let's not make the quick mistakes in questions like this, to use $n\rightarrow \infty$ and raise the power to $0$ getting the answer as $1$ as the digits in the brackets don't give a finite value.

The next step must be to try and bring the infinite terms in the denominators, to turn the complete value into $0$ . So, first of all, let's take something common out of the parenthesis.

$\quad \quad \quad \lim _{ n\rightarrow \infty }{ { ({ 4 }^{ n }+{ 8 }^{ n }) }^{ \frac { 1 }{ n } } } \\ \Rightarrow \quad \lim _{ n\rightarrow \infty }{ 8{ (1+\frac { 1 }{ { 2 }^{ n } } ) }^{ \frac { 1 }{ n } } } \\ \Rightarrow \quad 8\times \lim _{ n\rightarrow \infty }{ { (1+\frac { 1 }{ { 2 }^{ n } } ) }^{ \frac { 1 }{ n } } }$

Now, using $n\rightarrow \infty$ in the brackets, we get $\frac { 1 }{ n } \rightarrow 0\quad$ and $\quad \frac { 1 }{ { 2 }^{ n } } \rightarrow 0\quad$

Using these values, we can calculate the limit to be an easy $8$ !

Cheers!!:):)

Let's do this in a more general way: We will consider that $a>b$ :

So $a>b \Rightarrow \frac{b}{a}<1$

$\displaystyle \lim_{n\rightarrow \infty} (a^n+b^n)^{\frac{1}{n}}$

$\displaystyle \lim_{n\rightarrow \infty} a(1+(\frac{b}{a})^n)^{\frac{1}{n}}$

$\displaystyle \lim_{n\rightarrow \infty} e^{ln(a(1+(\frac{b}{a})^n)^{\frac{1}{n}})}$

$\displaystyle \lim_{n\rightarrow \infty} e^{ln(a)+\frac{ln(1+(\frac{b}{a})^n)}{n}}$

$\displaystyle \lim_{n\rightarrow \infty} (\frac{b}{a})^{n}=0 \Rightarrow \displaystyle \lim_{n\rightarrow \infty} \frac{ln(1+(\frac{b}{a})^n)}{n} =0$ Now we are left with the following limit:

$\displaystyle \lim_{n\rightarrow \infty} e^{ln(a)}=a$

And putting $a=8$ (because $8>4$ ), we get: $\boxed{\beta=8}$

Taking log

$\log y = \dfrac{1}{n}×\log(4^n+8^n)$

Using L hospital and then dividing the result obtained by 8^n, we get the answer as 8.