JEE Limits 2

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:

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

Thus, $\displaystyle \lim_{n\rightarrow \infty} (a^n+b^n)^{\frac{1}{n}}=a$ for $a>b$ .

Now, we can see the above limit turns out to be:

$\displaystyle \lim_{m\rightarrow \infty} \frac{1+2+3+...+m}{m^2}$

And: $1+2+3+...+m=\frac{m(m+1)}{2}$

So: $\displaystyle \lim_{m\rightarrow \infty} \frac{m(m+1)}{2m^2} =1/2$

Nice problem!

$\sqrt[n]{{(a-1)}^{n} + {a}^{n}}$

We will use Binomial expansion of $(a-1)^{n}$ .

$(a-1)^{n} = {a}^{n} - C(n,1){a}^{n-1} + C(n,2){a}^{n-2} -..... - {1}^{n}$

Now, we have to find $\displaystyle\lim_{n \rightarrow \infty}({a}^{n} - C(n,1){a}^{n-1} + C(n,2){a}^{n-2} -..... - {1}^{n})$

As $n \rightarrow\infty$ , $n \rightarrow n-1 \rightarrow n-2 \rightarrow.....$ . Hence, it becomes

$\displaystyle\lim_{n \rightarrow\infty}({a}^{n} - {1}^{n} + {a}^{n}(- C(n,1) + C(n,2) -..... ))$

We know that $-C(n,1) + C(n,2) -.... = 0$ . So, the limit becomes

$\displaystyle\lim_{n \rightarrow\infty}({a}^{n} - {1}^{n})$

The final limit will be $\displaystyle\lim_{n \rightarrow\infty}({({a}^{n} - {1}^{n})}^{\frac{1}{n}})$

which is easily equal to $a$ .

As a result, the limiting sum becomes $1 + 2 + 3 + 4 + ..... + (m-1) + m = \frac{m(m+1)}{2}$

$\displaystyle\lim_{n \rightarrow\infty}(\frac{m(m+1)}{2{m}^{2}}) = \frac{1}{2}$

Just take bigger number common from each bracket . as limit is towards infinity numbers less than 1 will go to zero

doing this you get get

1+2+3......m in numerator .

limit becomes (m+1)/2m which is 0.5 if m approaches infinity