Question

ans this question guys pls help me to solve this problem​

Answer 1

The above question can be solved by two methods

Method 1):-Solving using L'Hopital rule

Diffrentiate the given term with respect to.x

$\frac{dy}{dx} = \frac{2cos2x}{0 - \frac{1}{2} {(4 - x)}^{ \frac{ - 1}{2} } \times ( - 1) } \\ now \: put \: x = 0 \\ required \: limit = \frac{2cos0}{ \frac{ - 1}{2} { (\frac{1}{4}) }^{ \frac{1}{2} } \times - 1} = 8 \\ this \: will \: be \: the \: limit \\ \:$

Method2):-Solving using the formula of limits

divide the numerator by 2x and then multiply it by 2x,,so that the value doesn't effected

$= > \frac{ \frac{sin2x}{2x} \times 2x}{2 - \sqrt{4 - x} } \\ now \: we \: know \: that \: \frac{sinx}{x } = 1 \\ so \: \frac{sin2x}{2x} = 1 \\ = > \frac{2x \times 1}{2 - \sqrt{4 - x} } \\ multiply \: numerator \: and \: denominator \: by \\ 2 + \sqrt{4 - x} \\ = > \frac{2x(2 + \sqrt{4 - x}) }{(2 - \sqrt{4 - x})(2 + \sqrt{4 - x)} } \\ = > \frac{2x(2 + \sqrt{4 - x} )}{4 - 4 + x} = \frac{2(2 + \sqrt{4 - x}) }{1} \\ now \: put \: x = 0 \\ on \: putting \: we \: get \\ required \: limit = \frac{2(2 + 2)}{1} = 2 \times 4 = 8$

{hope it helps you dude}