Let
$f (x) = \frac{tan [e^{2}] x^{3} - tan [- e^{2}] x^{3}}{{sin}^{3} x}, x \neq 0$ , where $[x]$ is greatest integer function
The value of $f (0)$ for which f(x) is continuous is

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Solution

The correct option is B 15
Since $7 < e^{2} < 8$ , so $[e^{2}] = 7$ and $[- e^{2}] = - 8$
So $f (0) = lim x \to 0 f (x) = lim x \to 0 \frac{tan 7 x^{3} - tan (- 8) x^{3}}{{sin}^{3} x}$
$= lim x \to 0 (7 \frac{tan 7 x^{3}}{7 x^{3}} \frac{x^{3}}{{sin}^{3} x} + 8 \frac{tan 7 x^{3}}{7 x^{3}} \frac{x^{3}}{{sin}^{3} x})$
$= 7 + 8 = 15$

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Let f(x)=tan[e2]x3−tan[−e2]x3sin3x,x≠0, where [x] is greatest integer functionThe value of f(0) for which f(x) is continuous is

The correct option is B 15Since 7<e2<8, so [e2]=7 and [−e2]=−8So f(0)=limx→0f(x)=limx→0tan7x3−tan(−8)x3sin3x=limx→0(7tan7x37x3x3sin3x+8tan7x37x3x3sin3x)=7+8=15

Let
$f (x) = \frac{tan [e^{2}] x^{3} - tan [- e^{2}] x^{3}}{{sin}^{3} x}, x \neq 0$ , where $[x]$ is greatest integer function
The value of $f (0)$ for which f(x) is continuous is