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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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Plabita

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