Vtyjkyuk

$ fracsqrtn+1sqrtn$

Idk how to do this but my attempt:

$forall epsilon > 0, exists N > 0$, such that for all $n in mathbb N$, if $n > N$, then $left| fracsqrtn+1sqrtn - 1 right| < epsilon$

Let $epsilon > 0$ be arbitrary

Choose $N = $ _____ > 0

Suppose $n > N$, then $$= left|fracsqrtn+1 - sqrtnsqrtnright|$$

I don't know what to go from here

Let $epsilon>0$ and $N=frac12epsilon$.

Thus, for all $n>N$ we have $$sqrtnleft(sqrtn+1+sqrtnright)>2n>frac1epsilon,$$ which gives
$$sqrtnleft(sqrtn+1+sqrtnright)>frac1epsilon$$ or
$$frac1sqrtnleft(sqrtn+1+sqrtnright)<epsilon$$ or
$$fracsqrtn+1-sqrtnsqrtn<epsilon$$ or
$$left|sqrtfracn+1n-1right|<epsilon,$$
which says that our sequence converges to $1$.

Probably the best way to go forward is to multiply the expression by $$fracsqrtn+1+sqrt nsqrtn+1+sqrtn$$

Or divide by $sqrtn $ on top and bottom. .. get $frac sqrt1+frac1n1 $, which goes to $frac sqrt 1+01$, or $1$. So you may choose N large enough that $bigg|sqrt1+frac1N-1bigg|< epsilon $.Then do a little algebra to get how large N has to be in terms of $epsilon$... Get $N>frac 12epsilon +epsilon^2$.