JEE Mathematics 2023 Question with Solution

Given:

$$L=\lim_{n \to \infty} \frac{1 + 2 - 3 + 4 + 5 - 6 + \ldots + (3n - 2) + (3n - 1) - 3n}{\sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4}}$$

Find: The value of the limit.

Group the numerator in blocks of three terms:

$$(1+2-3)+(4+5-6)+\cdots+[(3n-2)+(3n-1)-3n]$$

Each block equals

$$(3k-2)+(3k-1)-3k=3k-3=3(k-1)$$

for $k=1,2,\ldots,n$. Hence the numerator becomes

$$0+3+6+\cdots+3(n-1)=3\left(0+1+2+\cdots+(n-1)\right)=\frac{3n(n-1)}{2}$$

Now the limit is

$$L=\lim_{n \to \infty} \frac{\frac{3n(n-1)}{2}}{\sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4}}$$

Rationalize the denominator:

$$\sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4} = \frac{(2n^4 + 4n + 3)-(n^4 + 5n + 4)}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}$$

so

$$\sqrt{2n^4 + 4n + 3} - \sqrt{n^4 + 5n + 4} = \frac{n^4-n-1}{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}$$

Therefore,

$$L=\lim_{n \to \infty} \frac{3n(n-1)}{2} \cdot \frac{\sqrt{2n^4 + 4n + 3} + \sqrt{n^4 + 5n + 4}}{n^4-n-1}$$

Factor $n^2$ from each square root:

$$\sqrt{2n^4 + 4n + 3}=n^2\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}, \qquad \sqrt{n^4 + 5n + 4}=n^2\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}$$

Hence,

$$L=\lim_{n \to \infty} \frac{3n(n-1)}{2} \cdot \frac{n^2\left(\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right)}{n^4-n-1}$$

As $n \to \infty$,

$$\frac{n(n-1)}{n^2} \to 1, \qquad \frac{n^4}{n^4-n-1} \to 1,$$

and

$$\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}} \to \sqrt{2}+1$$

Thus,

$$L=\frac{3}{2}(\sqrt{2}+1)$$

Therefore, the correct option is C.

The solution also contains a conflicting header saying option B, but the worked result is $\frac{3}{2}(\sqrt{2}+1)$, which matches option C.

Given: The same limit. Find: The quickest evaluation.

Observe directly that the numerator is a sum of block values:

$$(1+2-3),(4+5-6),\ldots,((3n-2)+(3n-1)-3n)$$

These are

$$0,3,6,\ldots,3(n-1)$$

so the numerator is

$$3\cdot \frac{(n-1)n}{2}$$

For the denominator, after rationalization its size is controlled by

$$\frac{n^4}{n^2(\sqrt{2}+1)}=\frac{n^2}{\sqrt{2}+1}$$

Hence the whole expression behaves like

$$\frac{\frac{3n(n-1)}{2}}{\frac{n^2}{\sqrt{2}+1}} \to \frac{3}{2}(\sqrt{2}+1)$$

Therefore, the correct option is C.