Question

Помогите вычислить предел

Answer 1

Ответ:

$4$

Пошаговое объяснение:

$\lim_{x \to 0} \dfrac{tg2x}{\sqrt{4+x}-\sqrt{4-x}}=\dfrac{tg(2 \cdot 0)}{\sqrt{4+0}-\sqrt{4-0}}=\dfrac{tg0}{\sqrt{4}-\sqrt{4}}=\dfrac{0}{0};$

Найдём предел по правилу Лопиталя:

$\lim_{x \to 0} \dfrac{tg2x}{\sqrt{4+x}-\sqrt{4-x}}=\lim_{x \to 0} \dfrac{(tg2x)'}{(\sqrt{4+x}-\sqrt{4-x})'}=$

$=\lim_{x \to 0} \dfrac{(2x)' \cdot \dfrac{1}{cos^{2}x}}{(\sqrt{4+x})'-(\sqrt{4-x})'}=\lim_{x \to 0} \dfrac{2 \cdot \dfrac{1}{cos^{2}x}}{\dfrac{1}{2\sqrt{4+x}} \cdot (4+x)'-\dfrac{1}{2\sqrt{4-x}} \cdot (4-x)'}=$$=\lim_{x \to 0} \dfrac{2}{cos^{2}x \bigg (\dfrac{1}{2\sqrt{4+x}}-\dfrac{1}{2\sqrt{4-x}} \cdot (-1) \bigg )}=$

$=\lim_{x \to 0} \dfrac{2}{cos^{2}x \bigg (\dfrac{1}{2\sqrt{4+x}}+\dfrac{1}{2\sqrt{4-x}} \bigg )}=\dfrac{2}{cos^{2}(0) \cdot \bigg (\dfrac{1}{2\sqrt{4+0}}+\dfrac{1}{2\sqrt{4-0}} \bigg )}=$

$=\dfrac{2}{1 \cdot \bigg (\dfrac{1}{2\sqrt{4}}+\dfrac{1}{2\sqrt{4}} \bigg )}=\dfrac{2}{\dfrac{1}{2 \cdot 2}+\dfrac{1}{2 \cdot 2}}=\dfrac{2}{\dfrac{1}{2 \cdot 2} \cdot 2}=\dfrac{1}{\dfrac{1}{4}}=\dfrac{4}{1}=4;$