The value of \mathop \lim \limits_n \to \infty 1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n \over \sqrt 2n^4 + 4n + 3 - \sqrt n^4 + 5n + 4 is :

Limits Continuity and Differentiability PYQ — JEE Main 2023 · 25 Jan · Shift 1 · Q28 with Solution

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Correct answer: B

Simplify the numerator

The numerator is $1+2-3+4+5-6+\cdots +(3n-2)+(3n-1)-3n.$

Group the terms in blocks of three: $[(1+2-3)] + [(4+5-6)] + \cdots + [(3n-2)+(3n-1)-3n].$

For the $k$ -th block, $(3k-2)+(3k-1)-3k = 3k-3 = 3(k-1).$

So the numerator becomes $\sum_{k=1}^{n} 3(k-1)=3\sum_{k=1}^{n}(k-1)=3\sum_{j=0}^{n-1} j.$

Now, $\sum_{j=0}^{n-1} j = \frac{(n-1)n}{2}.$

Hence numerator $=3\cdot \frac{n(n-1)}{2}=\frac{3n(n-1)}{2}.$

Simplify the denominator

The denominator is $\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}.$

To handle the limit, factor out $n^2$ from each square root: $\sqrt{2n^4+4n+3}=n^2\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}},$ $\sqrt{n^4+5n+4}=n^2\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}.$

Thus denominator $=n^2\left(\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right).$

As $n\to\infty$ , $\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}\to \sqrt{2}, \,\,\, \sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\to 1.$

So denominator behaves like $n^2(\sqrt{2}-1).$

Form the limit

Therefore,

=\lim_{n\to\infty} \frac{\frac{3n(n-1)}{2}}{n^2\left(\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right)}.$$ Divide numerator and denominator by $n^2$: $$=\lim_{n\to\infty} \frac{\frac{3}{2}\left(1-\frac{1}{n}\right)}{\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}}.$$ Now let $n\to\infty$: $$=\frac{3/2}{\sqrt{2}-1}.$$ Rationalize: $$\frac{3/2}{\sqrt{2}-1}=\frac{3}{2(\sqrt{2}-1)} =\frac{3(\sqrt{2}+1)}{2[(\sqrt{2}-1)(\sqrt{2}+1)]} =\frac{3(\sqrt{2}+1)}{2(2-1)}.$$ Hence, $$\boxed{\frac{3}{2}(\sqrt{2}+1)}.$$ --- 4. **Check options** - A: $\dfrac{3}{2\sqrt2}$ — incorrect - B: $\dfrac{3}{2}(\sqrt2+1)$ — correct - C: $3(\sqrt2+1)$ — incorrect - D: $\dfrac{\sqrt2+1}{2}$ — incorrect So the correct option is **B**.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: B

Simplify the numerator

The numerator is $1+2-3+4+5-6+\cdots +(3n-2)+(3n-1)-3n.$

Group the terms in blocks of three: $[(1+2-3)] + [(4+5-6)] + \cdots + [(3n-2)+(3n-1)-3n].$

For the $k$ -th block, $(3k-2)+(3k-1)-3k = 3k-3 = 3(k-1).$

So the numerator becomes $\sum_{k=1}^{n} 3(k-1)=3\sum_{k=1}^{n}(k-1)=3\sum_{j=0}^{n-1} j.$

Now, $\sum_{j=0}^{n-1} j = \frac{(n-1)n}{2}.$

Hence numerator $=3\cdot \frac{n(n-1)}{2}=\frac{3n(n-1)}{2}.$

Simplify the denominator

The denominator is $\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}.$

To handle the limit, factor out $n^2$ from each square root: $\sqrt{2n^4+4n+3}=n^2\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}},$ $\sqrt{n^4+5n+4}=n^2\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}.$

Thus denominator $=n^2\left(\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right).$

As $n\to\infty$ , $\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}\to \sqrt{2}, \,\,\, \sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\to 1.$

So denominator behaves like $n^2(\sqrt{2}-1).$

Form the limit

Therefore,

=\lim_{n\to\infty} \frac{\frac{3n(n-1)}{2}}{n^2\left(\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right)}.$$ Divide numerator and denominator by $n^2$: $$=\lim_{n\to\infty} \frac{\frac{3}{2}\left(1-\frac{1}{n}\right)}{\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}}.$$ Now let $n\to\infty$: $$=\frac{3/2}{\sqrt{2}-1}.$$ Rationalize: $$\frac{3/2}{\sqrt{2}-1}=\frac{3}{2(\sqrt{2}-1)} =\frac{3(\sqrt{2}+1)}{2[(\sqrt{2}-1)(\sqrt{2}+1)]} =\frac{3(\sqrt{2}+1)}{2(2-1)}.$$ Hence, $$\boxed{\frac{3}{2}(\sqrt{2}+1)}.$$ --- 4. **Check options** - A: $\dfrac{3}{2\sqrt2}$ — incorrect - B: $\dfrac{3}{2}(\sqrt2+1)$ — correct - C: $3(\sqrt2+1)$ — incorrect - D: $\dfrac{\sqrt2+1}{2}$ — incorrect So the correct option is **B**.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

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