Math Doubts

Power Rule of Limits


$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \Big(f(x)\Big)^{\displaystyle n} } \normalsize \,=\, \Big(\displaystyle \large \lim_{x \,\to\, a}{\normalsize f(x)} \normalsize \Big)^{\normalsize \displaystyle n}$

The limit of power of a function is equal to the power of limit of the function. It is called the power rule of limits.


$x$ is a variable and $f(x)$ is a function in terms of $x$. The literals $n$ and $a$ are constants. The function $f(x)$ and a constant $n$ formed a power function $\Big(f(x)\Big)^{\displaystyle n}$. The limit of the power function $\Big(f(x)\Big)^{\displaystyle n}$ as the input $x$ approaches $a$ is written in mathematics in the following form.

$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \Big(f(x)\Big)^{\displaystyle n}}$

The limit of $n$-th power of function $f(x)$ as $x$ tends to $a$ is equal to the $n$-th power of the limit of the function $f(x)$.

$\implies$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \Big(f(x)\Big)^{\displaystyle n} } \normalsize \,=\, \Big(\displaystyle \large \lim_{x \,\to\, a}{\normalsize f(x)} \normalsize \Big)^{\normalsize \displaystyle n}$


$(1) \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(x+6)}^2}$ $\,=\,$ $\Bigg[\displaystyle \large \lim_{x \,\to\, 0} \, {\normalsize {(x+6)}\Bigg]}^2$

$(2) \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize {(x^2+3x+4)}^5}$ $\,=\,$ $\Bigg[\displaystyle \large \lim_{x \,\to\, 2} \, {\normalsize {(x^2+3x+4)}\Bigg]}^5$

$(3) \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize {\Bigg(7+\dfrac{2}{x}\Bigg)}^{123}}$ $\,=\,$ $\Bigg[\displaystyle \large \lim_{x \,\to\, \infty} \, {\normalsize {\Bigg(7+\dfrac{2}{x}\Bigg)}\Bigg]}^{123}$


Learn how to derive the power rule of limits in mathematical form in calculus.

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