MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = $
- A$\frac{{{a^2} - {b^2}}}{2}$
- ✓$\frac{{{b^2} - {a^2}}}{2}$
- C${a^2} - {b^2}$
- D${b^2} - {a^2}$
$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{2\,\sin \,\left( {\frac{{a + b}}{2}} \right)x\,.\,\sin \,\left( {\frac{{b - a}}{2}} \right)\,x}}{{\left( {\frac{{a + b}}{2}} \right)x\,.\frac{2}{{a + b}}.\frac{2}{{b - a}}.\left( {\frac{{b - a}}{2}} \right)x}} = \frac{{{b^2} - {a^2}}}{2}$
Aliter : Apply $ L$-Hospital’s rule,
$\mathop {\lim }\limits_{x \to 0} \,\frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{ - \,a\sin ax + b\sin bx}}{{2x}}$
$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \,{a^2}\cos ax + {b^2}\cos bx}}{2} = \frac{{{b^2} - {a^2}}}{2}.$
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$a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}, n \geq 1$
$b_1=1 \text { and } b_n=a_{n-1}+a_{n+1}, n \geq 2.$
Then which of the following options is/are correct?
$(1)$ $a_1+a_2+a_3+\ldots . .+a_n=a_{n+2}-1$ for all $n \geq 1$
$(2)$ $\sum_{n=1}^{\infty} \frac{ a _{ n }}{10^{ n }}=\frac{10}{89}$
$(3)$ $\sum_{n=1}^{\infty} \frac{b_n}{10^n}=\frac{8}{89}$
$(4)$ $b=\alpha^n+\beta^n$ for all $n>1$