If \mathop \lim \limits_x \to 0 \ 1 \over x^8( 1 - \cos x^2 \over 2 - \cos x^2 \over 4 + \cos x^2 \over 2\cos x^2 \over 4 ) \ = 2-k then the value of k is \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2020 · 3 Sep · Shift 1 · Q39 with Solution

View written solutionFree

Correct answer: $\DISPLAYSTYLE \FRAC{511}{256}$

Let $A=\frac{x^2}{2},\qquad B=\frac{x^2}{4}.$ Then the given limit becomes $\lim_{x\to 0}\frac{1-\cos A-\cos B+\cos A\cos B}{x^8}.$
Notice that the numerator can be factorized: $1-\cos A-\cos B+\cos A\cos B=(1-\cos A)(1-\cos B).$ So the limit is $\lim_{x\to 0}\frac{(1-\cos A)(1-\cos B)}{x^8}.$
Use the standard small-angle expansion: $1-\cos t \sim \frac{t^2}{2} \quad (t\to 0).$ Hence, $1-\cos A \sim \frac{A^2}{2},\qquad 1-\cos B \sim \frac{B^2}{2}.$ Therefore, $(1-\cos A)(1-\cos B) \sim \frac{A^2B^2}{4}.$
Now compute $A^2$ and $B^2$ : $A^2=\left(\frac{x^2}{2}\right)^2=\frac{x^4}{4},$ $B^2=\left(\frac{x^2}{4}\right)^2=\frac{x^4}{16}.$ So, $A^2B^2=\frac{x^8}{64}.$ Hence, $(1-\cos A)(1-\cos B) \sim \frac{1}{4}\cdot \frac{x^8}{64}=\frac{x^8}{256}.$
Therefore,

=\frac{1}{256}.$$ 6. Given that this limit equals $2-k$, we get $$2-k=\frac{1}{256}.$$ So, $$k=2-\frac{1}{256}=\frac{511}{256}.$$ 7. Since this is an integer-type question, the stored answer $8$ does not match the evaluated expression. The mathematically derived value is $$k=\frac{511}{256}.$$

View written solutionFree

Correct answer: $\DISPLAYSTYLE \FRAC{511}{256}$

Let $A=\frac{x^2}{2},\qquad B=\frac{x^2}{4}.$ Then the given limit becomes $\lim_{x\to 0}\frac{1-\cos A-\cos B+\cos A\cos B}{x^8}.$
Notice that the numerator can be factorized: $1-\cos A-\cos B+\cos A\cos B=(1-\cos A)(1-\cos B).$ So the limit is $\lim_{x\to 0}\frac{(1-\cos A)(1-\cos B)}{x^8}.$
Use the standard small-angle expansion: $1-\cos t \sim \frac{t^2}{2} \quad (t\to 0).$ Hence, $1-\cos A \sim \frac{A^2}{2},\qquad 1-\cos B \sim \frac{B^2}{2}.$ Therefore, $(1-\cos A)(1-\cos B) \sim \frac{A^2B^2}{4}.$
Now compute $A^2$ and $B^2$ : $A^2=\left(\frac{x^2}{2}\right)^2=\frac{x^4}{4},$ $B^2=\left(\frac{x^2}{4}\right)^2=\frac{x^4}{16}.$ So, $A^2B^2=\frac{x^8}{64}.$ Hence, $(1-\cos A)(1-\cos B) \sim \frac{1}{4}\cdot \frac{x^8}{64}=\frac{x^8}{256}.$
Therefore,

=\frac{1}{256}.$$ 6. Given that this limit equals $2-k$, we get $$2-k=\frac{1}{256}.$$ So, $$k=2-\frac{1}{256}=\frac{511}{256}.$$ 7. Since this is an integer-type question, the stored answer $8$ does not match the evaluated expression. The mathematically derived value is $$k=\frac{511}{256}.$$