In this limit problem, there are three terms in radical form in the numerator. The limit of the algebraic function can be evaluated by simplifying this function same as the known form.
$\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{\sqrt[3]{x}+\sqrt[4]{x}+\sqrt[5]{x}-3}{x-1}}$
In numerator, the first three terms have different radical symbols and they can be expressed in exponential form.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}+x^\frac{1}{4}+x^\frac{1}{5}-3}{x-1}}$
According to limit rule of algebraic function, the limit of algebraic function can be converted same as the limit of $\dfrac{x^n-a^n}{x-a}$ as $x$ approaches $a$ rule.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}+x^\frac{1}{4}+x^\frac{1}{5}-1-1-1}{x-1}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}-1+x^\frac{1}{4}-1+x^\frac{1}{5}-1}{x-1}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \Bigg[\dfrac{x^\frac{1}{3}-1}{x-1}+\dfrac{x^\frac{1}{4}-1}{x-1}+\dfrac{x^\frac{1}{5}-1}{x-1}\Bigg]}$
Now, use addition rule of limits to express limit of sum of three functions as sum of their limits.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}-1}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{4}-1}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{5}-1}{x-1}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}-{(1)}^\frac{1}{3}}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{4}-{(1)}^\frac{1}{4}}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{5}-{(1)}^\frac{1}{5}}{x-1}}$
$\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}-{(1)}^\frac{1}{3}}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{4}-{(1)}^\frac{1}{4}}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{5}-{(1)}^\frac{1}{5}}{x-1}}$
Now, evaluate limit of each function as per limit of (xn-an)/(x-a) as x approaches a rule.
$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{1}{3}-1}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{1}{4}-1}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{1}{5}-1}$
$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{1-3}{3}}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{1-4}{4}}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{1-5}{5}}$
$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{-2}{3}}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{-3}{4}}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{-4}{5}}$
$= \,\,\,$ $\dfrac{1}{3} \times 1$ $+$ $\dfrac{1}{4} \times 1$ $+$ $\dfrac{1}{5} \times 1$
$= \,\,\,$ $\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}$
$= \,\,\,$ $\dfrac{20 \times 1 + 15 \times 1 + 12 \times 1}{60}$
$= \,\,\,$ $\dfrac{20+15+12}{60}$
$= \,\,\,$ $\dfrac{47}{60}$
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