The value of \lim_x arrow 0 2(\frac1-\cos x \sqrt\cos 2 x \sqrt[3]\cos 3 x \ldots \ldots . \sqrt[10]\cos 10 xx^2) is \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q60 with Solution

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Correct answer: 55

Let $P(x)=ig(\cos x\big)\big(\cos 2x\big)^{1/2}\big(\cos 3x\big)^{1/3}\cdots \big(\cos 10x\big)^{1/10}.$

We need $\lim_{x\to 0}2\left(\frac{1-P(x)}{x^2}\right).$

Since $x\to 0$ , each factor tends to $1$ . For such products, taking logarithm is convenient: $\ln P(x)=\sum_{k=1}^{10}\frac{1}{k}\ln(\cos kx).$
Use the standard expansion near $0$ : $\cos u = 1-\frac{u^2}{2}+O(u^4),$ so $\ln(\cos u)= -\frac{u^2}{2}+O(u^4).$

Substitute $u=kx$ : $\ln(\cos kx)= -\frac{k^2x^2}{2}+O(x^4).$

Therefore,

= -\frac{x^2}{2}\sum_{k=1}^{10}k + O(x^4).$$ Now, $$\sum_{k=1}^{10}k=\frac{10\cdot 11}{2}=55.$$ Hence, $$\ln P(x)= -\frac{55}{2}x^2+O(x^4).$$ 4. Exponentiating, $$P(x)=e^{\ln P(x)}=e^{-\frac{55}{2}x^2+O(x^4)}=1-\frac{55}{2}x^2+O(x^4).$$ So, $$1-P(x)=\frac{55}{2}x^2+O(x^4).$$ 5. Therefore, $$2\left(\frac{1-P(x)}{x^2}\right)=2\left(\frac{\frac{55}{2}x^2+O(x^4)}{x^2}\right)=55+O(x^2).$$ Thus, $$\lim_{x\to 0}2\left(\frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}\cdots \sqrt[10]{\cos 10x}}{x^2}\right)=55.$$ 6. Comparing with the stored correct answer: stored answer is $55$, which matches the derived result.

View written solutionFree

Correct answer: 55

Let $P(x)=ig(\cos x\big)\big(\cos 2x\big)^{1/2}\big(\cos 3x\big)^{1/3}\cdots \big(\cos 10x\big)^{1/10}.$

We need $\lim_{x\to 0}2\left(\frac{1-P(x)}{x^2}\right).$

Since $x\to 0$ , each factor tends to $1$ . For such products, taking logarithm is convenient: $\ln P(x)=\sum_{k=1}^{10}\frac{1}{k}\ln(\cos kx).$
Use the standard expansion near $0$ : $\cos u = 1-\frac{u^2}{2}+O(u^4),$ so $\ln(\cos u)= -\frac{u^2}{2}+O(u^4).$

Substitute $u=kx$ : $\ln(\cos kx)= -\frac{k^2x^2}{2}+O(x^4).$

Therefore,

= -\frac{x^2}{2}\sum_{k=1}^{10}k + O(x^4).$$ Now, $$\sum_{k=1}^{10}k=\frac{10\cdot 11}{2}=55.$$ Hence, $$\ln P(x)= -\frac{55}{2}x^2+O(x^4).$$ 4. Exponentiating, $$P(x)=e^{\ln P(x)}=e^{-\frac{55}{2}x^2+O(x^4)}=1-\frac{55}{2}x^2+O(x^4).$$ So, $$1-P(x)=\frac{55}{2}x^2+O(x^4).$$ 5. Therefore, $$2\left(\frac{1-P(x)}{x^2}\right)=2\left(\frac{\frac{55}{2}x^2+O(x^4)}{x^2}\right)=55+O(x^2).$$ Thus, $$\lim_{x\to 0}2\left(\frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}\cdots \sqrt[10]{\cos 10x}}{x^2}\right)=55.$$ 6. Comparing with the stored correct answer: stored answer is $55$, which matches the derived result.