Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{n\to \infty }\frac{1+2-3+4+5-6+...+(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$ is :

Finding Order in Chaos

The numerator is given by the expression $1+2-3+4+5-6+...+(3n-2)+(3n-1)-3n$. At first glance, this appears to be a chaotic sequence, but there is a distinct rhythm where every third term is negative.

We group these terms into triplets. We define the $k$-th group as $G_k=(3k-2)+(3k-1)-3k$. Simplifying this, we find:

For $k=1$, the value is $0$. For $k=2$, the value is $3$. For $k=3$, the value is $6$. The sequence transforms into an arithmetic progression: $0+3+6+...+3(n-1)$.

Using the sum formula for an arithmetic progression, $S_n=\frac{n}{2}(a+l)$, the sum of the numerator becomes:

The Power of Dominance

Now, we examine the denominator: $\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}$. As $n\to \infty$, the lower-order terms $4n,3,5n,$ and $4$ become negligible compared to the $n^4$ terms.

We factor out $n^4$ from inside the radicals, which pulls out as $n^2$:

As $n\to \infty$, the fractional terms approach zero. The expression simplifies to:

Final Calculation

We now combine the simplified numerator and denominator to evaluate the limit:

The $n^2$ terms cancel out, leaving us with the expression:

To reach the final answer, we rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{2}+1)$:

The final result is $\frac{3}{2}(\sqrt{2}+1)$.