limit without expansion I was solving this limit  instead of using sum of expansion i  did this and got zero  but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots +  n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n^5}$$
$$= \lim_{n\rightarrow \infty } \frac{( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n}$$
$$= \frac{(0 + 0 +0 \ldots + 1 )}{n} = 0$$ this is how i got  ,

You were going right till the very last step. You need to apply the limit after computing the sum. The limit converts to Riemann integral:
$$\lim_{n\to \infty}\sum_{k=1}^{n} \frac{k^4}{n^5} = \int_{0}^{1} x^4 dx = \frac{1}{5}$$

Since you are not familiar with integration, another method is to find $\sum_{k=1}^{n} k^4$ explicitly. Only the leading coefficient, ie the coefficient of $n^5$ needs to be found.
$$\sum_{k=1}^nk^4=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)$$
So we can write $\sum_{k=1}^nk^4 = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} -\frac{n}{30}$. Thus if you compute the limit
$$\lim_{n\to \infty}\sum_{k=1}^n\frac{k^4}{n^5} = \lim_{n\to\infty} \frac{\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} -\frac{n}{30}}{n^5}\\
\lim_{n\to\infty} \frac{\frac{1}{5} + \frac{1}{2n} + \frac{1}{3n^2} -\frac{1}{30n^4}}{1} = \frac{1}{5}$$
