CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

$$\mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{1}{{h\sqrt[3]{{8 + h}}}} - \dfrac{1}{{2h}}} \right]$$


Solution

$$\displaystyle \lim_{n \to 0} \left [\dfrac {1}{n \sqrt [2]{8+h}}-\dfrac {1}{2n}\right]$$
$$\displaystyle \lim_{n \to 0} \left [\dfrac {2-\sqrt [2]{8+h}}{h \sqrt [3]{8+h}}\right]$$
$$\displaystyle \lim_{n \to 0} \dfrac {(2-\sqrt [3]{8+h})(2)^2 + (\sqrt [2]{8+h})^2+2.\sqrt [3]{8+h}}{(h \sqrt [3]{8+h})(2)^2 + \sqrt [3]{8+h}+2\sqrt [3]{8+h}}$$
$$\displaystyle \lim_{n \to 0} \dfrac {(2)^3 - (8+h)}{n\sqrt [3]{8+h} (4+(\sqrt [3]{8+h})^2 +2\sqrt [3]{8+h})}$$
$$\displaystyle \lim_{n\to 0} \dfrac {-1}{(\sqrt [3]{8+h}) (4+(\sqrt [3]{8+h})^2 + 2\sqrt [3]{8+h})}$$
$$=\dfrac {-1}{2.(4+1+4)}$$
$$=\dfrac {-1}{24}$$Ans

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image