Menu Close

lim-x-a-x-1-m-a-1-m-x-1-n-a-1-n-Don-t-use-L-hospital-rules-

Tinku Tara 0 Comments
Pin




Question Number 29180 by A1B1C1D1 last updated on 05/Feb/18
lim_(x → a) ((((x)^(1/m) − (a)^(1/m) )/( (x)^(1/n) − (a)^(1/n) ))) Don′t use L′hospital rules :-)
$$\underset{\mathrm{x}\:\rightarrow\:\mathrm{a}} {\mathrm{lim}}\:\left(\frac{\sqrt[{\mathrm{m}}]{\mathrm{x}}\:−\:\sqrt[{\mathrm{m}}]{\mathrm{a}}}{\:\sqrt[{\mathrm{n}}]{\mathrm{x}}\:−\:\sqrt[{\mathrm{n}}]{\mathrm{a}}}\right) \\ $$$$\left.\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{L}'\mathrm{hospital}\:\mathrm{rules}\::-\right) \\ $$
Answered by Rasheed.Sindhi last updated on 05/Feb/18
Formula lim_(x→a) ((x^n −a^n )/(x−a))=na^(n−1) lim_(x → a) ((((x)^(1/m) − (a)^(1/m) )/( (x)^(1/n) − (a)^(1/n) )))=lim_(x→a) ((x^(1/m) −a^(1/n) )/(x^(1/n) −a^(1/n) )) =lim_(x→a) (( ((x^(1/m) −a^(1/n) )/(x−a)) )/((x^(1/n) −a^(1/n) )/(x−a))) =lim_(x→a) (( lim_(x→a) ((x^(1/m) −a^(1/n) )/(x−a)) )/(lim_(x→a) ((x^(1/n) −a^(1/n) )/(x−a))))=(( (1/m)a^((1/m)−1) )/((1/n)a^((1/n)−1) )) =(n/m)a^((1/m)−(1/n)) =(n/m)a^((n−m)/(nm))
$$\mathrm{Formula} \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{n}} −\mathrm{a}^{\mathrm{n}} }{\mathrm{x}−\mathrm{a}}=\mathrm{na}^{\mathrm{n}−\mathrm{1}} \\ $$$$\underset{\mathrm{x}\:\rightarrow\:\mathrm{a}} {\mathrm{lim}}\:\left(\frac{\sqrt[{\mathrm{m}}]{\mathrm{x}}\:−\:\sqrt[{\mathrm{m}}]{\mathrm{a}}}{\:\sqrt[{\mathrm{n}}]{\mathrm{x}}\:−\:\sqrt[{\mathrm{n}}]{\mathrm{a}}}\right)=\underset{\mathrm{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{1}/\mathrm{m}} −\mathrm{a}^{\mathrm{1}/\mathrm{n}} }{\mathrm{x}^{\mathrm{1}/\mathrm{n}} −\mathrm{a}^{\mathrm{1}/\mathrm{n}} } \\ $$$$\:\:\:\:\:=\underset{\mathrm{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\:\:\:\:\:\frac{\mathrm{x}^{\mathrm{1}/\mathrm{m}} −\mathrm{a}^{\mathrm{1}/\mathrm{n}} }{\mathrm{x}−\mathrm{a}}\:\:\:\:\:}{\frac{\mathrm{x}^{\mathrm{1}/\mathrm{n}} −\mathrm{a}^{\mathrm{1}/\mathrm{n}} }{\mathrm{x}−\mathrm{a}}} \\ $$$$\:\:\:\:\:=\underset{\mathrm{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{1}/\mathrm{m}} −\mathrm{a}^{\mathrm{1}/\mathrm{n}} }{\mathrm{x}−\mathrm{a}}\:\:\:\:\:}{\underset{\mathrm{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{1}/\mathrm{n}} −\mathrm{a}^{\mathrm{1}/\mathrm{n}} }{\mathrm{x}−\mathrm{a}}}=\frac{\:\:\frac{\mathrm{1}}{\mathrm{m}}\mathrm{a}^{\frac{\mathrm{1}}{\mathrm{m}}−\mathrm{1}} }{\frac{\mathrm{1}}{\mathrm{n}}\mathrm{a}^{\frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{n}}{\mathrm{m}}\mathrm{a}^{\frac{\mathrm{1}}{\mathrm{m}}−\frac{\mathrm{1}}{\mathrm{n}}} =\frac{\mathrm{n}}{\mathrm{m}}\mathrm{a}^{\frac{\mathrm{n}−\mathrm{m}}{\mathrm{nm}}} \\ $$
Commented by A1B1C1D1 last updated on 05/Feb/18
Thanks
$$\mathrm{Thanks} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *