If $$\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\}$$ = 2-k
then the value of k is _______ .
Answer (integer)
8
Solution
$$\mathop {\lim }\limits_{x \to 0} \left\{ {{1 \over {{x^8}}}\left( {1 - \cos {{{x^2}} \over 2} - \cos {{{x^2}} \over 4} + \cos {{{x^2}} \over 2}\cos {{{x^2}} \over 4}} \right)} \right\} = {2^{ - k}}$$
<br><br>$\Rightarrow$ $$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos {{{x^2}} \over 2}} \right)} \over {4{{\left( {{{{x^2}} \over 2}} \right)}^2}}}{{\left( {1 - \cos {{{x^2}} \over 4}} \right)} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} $$ = ${2^{ - k}}$
<br><br>$\Rightarrow$ $$\mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}{{{x^2}} \over 4}} \over {16{{\left( {{{{x^2}} \over 4}} \right)}^2}}} \times {{2{{\sin }^2}{{{x^2}} \over 8}} \over {64{{\left( {{{{x^2}} \over 8}} \right)}^2}}}$$ = 2<sup>-k</sup>
<br><br>$\Rightarrow$ ${1 \over 8} \times {1 \over {32}} = {2^{ - k}}$
<br><br>$\Rightarrow$ 2<sup>-8</sup> = 2<sup>-k</sup>
<br><br>$\Rightarrow$ k = 8
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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