Question Number 92511 by Ar Brandon last updated on 07/May/20
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{x}}−\sqrt[{\mathrm{5}}]{\mathrm{x}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}−\sqrt[{\mathrm{4}}]{\mathrm{x}}} \\ $$
Commented by john santu last updated on 07/May/20
$$\mathrm{x}\:=\:\mathrm{t}^{\mathrm{60}} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{t}^{\mathrm{20}} −\mathrm{t}^{\mathrm{12}} }{\mathrm{t}^{\mathrm{20}} −\mathrm{t}^{\mathrm{15}} }\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{t}^{\mathrm{12}} \left(\mathrm{t}^{\mathrm{8}} −\mathrm{1}\right)}{\mathrm{t}^{\mathrm{15}} \left(\mathrm{t}^{\mathrm{5}} −\mathrm{1}\right)} \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{t}^{\mathrm{8}} −\mathrm{1}}{\mathrm{t}^{\mathrm{3}} \left(\mathrm{t}^{\mathrm{5}} −\mathrm{1}\right)}\:=\:\mathrm{DNE}\: \\ $$
Commented by Ar Brandon last updated on 07/May/20
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