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if $f(x)$ is continuous and $f(\frac{9}{2})=\frac{2}{9}$ , then $\lim_{x\rightarrow 0}f\left ( \frac{1-\cos 3x}{x^{2}} \right )$ is equal to :

Option 1)
$\frac{9}{2}$

Option 2)
$\frac{2}{9}$

Option 3)
0

Option 4)
$\frac{8}{9}$

Answers (1)

best_answer

As we learnt in

Evalution of Trigonometric limit -

$\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1$

$\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1$

$put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0$

$Then\:it\:comes$

$\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1$

-

$\lim_{x\rightarrow 0} \:f \left ( \frac{1-\cos 3x}{x^{2}} \right )$

$\lim_{x\rightarrow 0} \:f \left ( \frac{2 \sin^{2}\:\frac{3x}{2}}{x^{2}} \right )$

$\lim_{x\rightarrow 0} \:f \left ( \frac{2 \sin^{2}\:\frac{3x}{2}}{\frac{9x^{2}}{4}} \times \frac{9}{4} \right)$

$\Rightarrow \left ( \frac{9}{2} \right )= \frac{2}{9} \:\:\:\:\: (given)$

Option 1)

$\frac{9}{2}$

This option is incorrect.

Option 2)

$\frac{2}{9}$

This option is correct.

Option 3)

0

This option is incorrect.

Option 4)

$\frac{8}{9}$

This option is incorrect.

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