Question

Find the value of $\underset{x\to 0}{lim}\frac{\sqrt[K]{K1+x}-1}{x}$, where $K$ is a positive integer

• A. $K$
• B. $-K$
• C. $\frac{1}{K}$
• D. $-\frac{1}{K}$

Answer 1

The correct option is C $\frac{1}{K}$

When plugging $x=0$ we get $\frac{0}{0}$ indeterminate form so let's apply the L'Hospital Rule

${lim}_{x\to 0}\frac{f\left(x\right)}{g\left(x\right)}={lim}_{x\to 0}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

Let us assume that $f\left(x\right)=\sqrt[k]{k1+x}-1$ and $g\left(x\right)=x$

Differentiate both the function, we get

$f^{\prime }\left(x\right)=\frac{(x+1{)}^{\frac{1-k}{k}}}{k}$ and $g^{\prime }\left(x\right)=1$

So,

${lim}_{x\to 0}\frac{f^{\prime }\left(x\right))}{g^{\prime }\left(x\right)}$ $={lim}_{x\to 0}\frac{\frac{(x+1{)}^{\frac{1-k}{k}}}{k}}{1}$ $=\frac{(0+1{)}^{\frac{1-k}{k}}}{k}$ $=\frac{1}{k}$