Math, asked by Anonymous, 10 months ago

$\huge{\underline{\overline{\mathtt{\mid{Evaluate:-}\mid}}}}$
limit x tends to 0
$\huge \frac{ \ \cos2x - 1 }{ \cos \: x - 1 }$

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Answered by RvChaudharY50

179

Question :------

limit x → 0
$\huge \frac{ \ \cos2x - 1 }{ \cos \: x - 1 }$

Formula used :------

cos2x = 1 - 2 sin²x
sin2x = 2 sinx × cosx
cox0° = 1
L hospital rule ....

Solution :-----

According to L hospital rule we know that,,

limit x → 0

$\huge\frac{f(x)}{g(x)}$

= limit x → 0

$\huge\frac{f'(x)}{g'(x)}$

so,

limit x → 0

$\huge \frac{ \ \cos2x - 1 }{ \cos \: x - 1 }$

$\red{\boxed\implies} \: lim \: x \iff0 = \frac{ - sin2x}{ - sinx} \\ \\ \red{\boxed\implies} \: lim \: x \iff0 = \: \frac{ \cancel- 4 \cancel{sinx}cosx}{ \cancel{ - sinx}} \\ \\ \red{\boxed\implies} \: lim \: x \iff0 = \: 4cosx \\ \\ now \: since \: cosx \: at \: x \: = 0 \: is \: 1.. \\ \\ \\ \\ so \: \\ \\ \\ \red{\boxed\implies} \: lim \: x \iff0 = \: 4cosx \: = \red{\large\boxed{\bold{4}}}$

so, value of given date is 4 ....

(Hope it helps you) ...

Answered by Anonymous

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$\huge{\underline{\overline{\mathtt{\mid{Evaluate:-}\mid}}}}$
limit x tends to 0
$\huge \frac{ \ \cos2x - 1 }{ \cos \: x - 1 }$