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Evaluate $\displaystyle \Large \lim_{x \,\to\, 0} \, \normalsize \dfrac{\sqrt[\displaystyle k]{1+x}-1}{x}$

$x$ and $k$ are two literals. The division of subtraction of $1$ from the $k^{th}$ root of sum of $1$ and $x$ by $x$ formed an algebraic function.

The value of the function should be evaluated as $x$ approaches zero.

$\displaystyle \Large \lim_{x \,\to\, 0} \, \normalsize \dfrac{\sqrt[\displaystyle k]{1+x}-1}{x}$


Substitute x = 0

Put $x = 0$ in the algebraic function to find its value.

$=\,\,\,$ $\displaystyle \dfrac{\sqrt[\displaystyle k]{1+0}-1}{0}$

$=\,\,\,$ $\displaystyle \dfrac{\sqrt[\displaystyle k]{1}-1}{0}$

$=\,\,\,$ $\displaystyle \dfrac{1-1}{0}$

$=\,\,\,$ $\displaystyle \dfrac{0}{0}$

The value of the function is indeterminate as $x$ approaches zero. Hence, solve this limit problem in another method.


Transforming the function

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 0} \, \normalsize \dfrac{\sqrt[\displaystyle k]{x+1}-1}{x}$

If $x \to 0$, then $x+1 \to 1$

$=\,\,\,$ $\displaystyle \Large \lim_{x+1 \,\to\, 1} \, \normalsize \dfrac{\sqrt[\displaystyle k]{x+1}-1}{x}$

Take $h = x+1$, then $x = h-1$. Eliminate $x$ and transform the entire function in terms of $h$.

$=\,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 1} \, \normalsize \dfrac{\sqrt[\displaystyle k]{h}-1}{h-1}$

$=\,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 1} \, \normalsize \dfrac{{(h)}^\dfrac{1}{k}-1}{h-1}$

$=\,\,\,$ $\displaystyle \Large \lim_{h \,\to\, 1} \, \normalsize \dfrac{{(h)}^\dfrac{1}{k}-{(1)}^\dfrac{1}{k}}{h-1}$


Apply Limit formula

The function is simplified same as the following exponential limit formula.

$\displaystyle \Large \lim_{x \,\to\, a} \large \dfrac{x^n-a^n}{x-a} \,=\, n.a^{n-1}$

Use this formula and evaluate the value of the limit of the function.

$=\,\,\,$ $\dfrac{1}{k} {(1)}^{\dfrac{1}{k}\,-\,1}$

$=\,\,\,$ $\dfrac{1}{k} {(1)}^{\dfrac{1-k}{k}}$

$=\,\,\,$ $\dfrac{1}{k} \times 1$

$=\,\,\,$ $\dfrac{1}{k}$

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