MCQ
$\mathop {\lim }\limits_{m \to \infty } \,{\left( {\cos \frac{x}{m}} \right)^m} = $
- A$0$
- B$e$
- C$1/e$
- ✓$1$
$ = \mathop {\lim }\limits_{m \to \infty } {\left[ {1 - \left( { - \cos \frac{x}{m} + 1} \right)} \right]^m}$
$ = \mathop {\lim }\limits_{m \to \infty } {\left[ {1 - 2{{\sin }^2}\frac{x}{{2m}}} \right]^m}$
$ = {e^{\mathop {\lim }\limits_{m \to \infty } - \left( {2{{\sin }^2}\frac{x}{{2m}}} \right)\,m}}$
$ = {e^{\mathop {\lim }\limits_{m \to \infty } - 2{{\left( {\frac{{\sin \frac{x}{{2m}}}}{{x/2m}}} \right)}^2}\left( {\frac{{{x^2}}}{{4{m^2}}}} \right)\,m}}$
$ = {e^{ - 2\mathop {\lim }\limits_{m \to \infty } \frac{{{x^2}}}{{4m}}}} = {e^0} = 1$.
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Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1{ }^2-\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2-\mathrm{a}_4{ }^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}{ }^2-\mathrm{a}_{2 \mathrm{k}}{ }^2$.
If $\mathrm{A}_3=-153, \mathrm{~A}_5=-435$ and $\mathrm{a}_1{ }^2+\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2=66$, then $\mathrm{a}_{17}-\mathrm{A}_7$ is equal to....................