Question

ǫᴜᴇsᴛɪᴏɴ ( ᴊᴇᴇ ) ❤ :

sᴏʟᴠᴇ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ ᴏғ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ ^_^ .....

✮ɢɪᴠᴇ ᴅᴇᴛᴀɪʟᴇᴅ ᴇxᴘʟᴀɴᴀᴛɪᴏɴ✮

ᴛʜᴀɴᴋs :)
ᴀʟʟ ᴛʜᴇ ʙᴇsᴛ ❤ .

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Answer 1

EXPLANATION.

$\displaystyle\lim_{x \to \: - \infty } = ( \frac{x {}^{4} \sin( \frac{1}{x} ) + {x}^{2} }{(1 + |x| {}^{3} ) } )$

$it \: \: is \: \: in \: \: form \: \: of \: = \frac{ \infty }{ \infty }$

$\displaystyle\lim_{x \to \: - \infty } = ( \frac{x {}^{4} \sin( \frac{1}{x} ) + {x}^{2} }{1 - x {}^{3} } )$

Divide numerator and denominator by x³

$\displaystyle\lim_{x \to \: - \infty } = ( \frac{ \frac{x {}^{4} }{x {}^{3} } \sin( \frac{1}{x} ) + \frac{x {}^{2} }{x {}^{3} } }{ \frac{1}{x {}^{3} } - \frac{ {x}^{3} }{x {}^{3} } } )$

$\displaystyle\lim_{x \to \: - \infty } = ( \frac{x \sin( \frac{1}{x} ) + \frac{1}{x} }{ \frac{1}{x {}^{3} } - 1 } )$

Therefore,

we get,

$\displaystyle\lim_{x \to \: - \infty } = ( \frac{1 + 0}{0 - 1} )$

$\displaystyle\lim_{x \to \: - \infty } = - 1$

value = -1

Answer 2

Please refer to the attachment.

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- A JEE aspirant this side.