Question

please ans limit class 12th​

Answer 1

Solution:-To find limit

$(x - > 0)( \frac{cos \alpha - \sec \alpha }{ {x}^{2}(x + 1)})$

proof:=>

$\frac{cosx - {cos}^{ - }x}{ {x}^{2}(x + 1) } \\ \: using \: l \: hopital \: rule \: \\ diffrentiate \: the \: term \: with \: respect \: to \: x \\ \frac{dy}{dx} = \frac{ - sinx + sinx \times cosx}{2x(x + 1) + {x}^{2} } \\ now \: putting \: x = 0 \: we \: again \: get \: \frac{0}{0} \\ as \: indiscriminant \: form \\ diffrentiate \: it \: again \\ = > \frac{ - cosx - {sin}^{2}x \: + {cos}^{2}x}{6x + 2} \\ now \: put \: x = 0 \\ = > \frac{ - 1 - 0 + 1}{0 + 2} = 0 \\ hence \: 0 \: is \: the \: required \: limit$