Question

$\underset{x\to 1}{lim}\frac{\sqrt[n]{nx-1}}{\sqrt[m]{mx-1}}$ (m and n integers) is equal to

• A. $0$
• B. $1$
• C. $\frac{m}{n}$
• D. $\frac{n}{m}$

Answer 1

The correct option is A $0$

We have,

$\underset{x\to 1}{lim}\frac{\sqrt[n]{nx-1}}{\sqrt[m]{mx-1}}$

This is $\frac{0}{0}$ form, so apply L-Hospital rule

$\underset{x\to 1}{lim}\frac{\frac{1}{n}(x-1{)}^{\frac{1-n}{n}}\times (1-0)}{\frac{1}{m}(x-1{)}^{\frac{1-m}{m}}\times (1-0)}$

$\underset{x\to 1}{lim}\frac{m}{n}(x-1{)}^{\frac{1-n}{n}-\frac{1-m}{m}}$

$\underset{x\to 1}{lim}\frac{m}{n}(x-1{)}^{\frac{m-mn-n+mn}{mn}}$

$\underset{x\to 1}{lim}\frac{m}{n}(x-1{)}^{\left(\frac{m-n}{mn}\right)}$

$=\frac{m}{n}\times (1-1{)}^{\left(\frac{m-n}{mn}\right)}$

$=\frac{m}{n}\times 0=0$

Hence, this is the answer.