Question

solve the following limit please... i will mark u brainliest​

Answer 1

Answer:

  Apply L'hospital's rule and differentiate both numerator and denominator.

And then apply limit x=0.

Step-by-step explanation:

This will be the final answer

Answer 2

Explanation :

$\tt \bigstar \: lim_{x \to 0} \bigg( \cfrac{ \sqrt{3 + x} - \sqrt{3 - x} }{x {}^{3} + 4x } \bigg) \\ \\ \\ \underline{\tt \: Applying \: L - Hospital \: rule, }\\ \\ \\ : \implies \: \tt \: lim_{x \to 0} \bigg( \cfrac{ \cfrac{d}{dx}( \sqrt{3 + x} - \sqrt{3 - x})}{ \cfrac{d}{dx} (x {}^{3} + 4x)} \bigg) \\ \\ \\ : \implies \: \tt \: lim_{x \to 0} \bigg( \cfrac{ \cfrac{1}{2 \sqrt{3 + x} } + \cfrac{1}{2 \sqrt{3 - x} } }{3x {}^{2} + 4 } \bigg) \\ \\ \\ : \implies \: \tt \: lim_{x \to 0} \bigg( \cfrac{ \cfrac{ \sqrt{3 - x} + \sqrt{3 + x} }{2 \sqrt{9 - x {}^{2} } } }{3 x{}^{2} + 4} \bigg) \\ \\ \\ : \implies \: \tt \: lim_{x \to 0} \bigg( \dfrac{ \sqrt{3 - x} + \sqrt{3 - x} }{2 \sqrt{9 - x {}^{2} }(3x {}^{2} + 4) } \bigg) \\ \\ \\ : \implies \: \tt \bigg( \cfrac{ \sqrt{3 - 0} + \sqrt{3 - 0} }{2 \sqrt{9 - 0 {}^{2} } (3 \times 0 {}^{2} + 4) } \bigg) \\ \\ \\ : \implies \: \tt \bigg( \cfrac{2 \sqrt{3} }{6(4)} \bigg) \\ \\ \\ : \implies \: \tt \bigg( \cfrac{ \not2 \sqrt{3} }{ \not{24}} \bigg) \\ \\ \\ : \implies { \underline{ \boxed{\tt \cfrac{ \sqrt{3} }{12} }}} \bigstar$

Hence :

$\dag{\boxed{\tt lim_{x \to 0} \bigg( \cfrac{ \sqrt{3 + x} - \sqrt{3 - x} }{x {}^{3} + 4x } \bigg) =\cfrac{\sqrt{3}}{12}}}$