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.
$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{1}{3}-1}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{1}{4}-1}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{1}{5}-1}$
$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{1-3}{3}}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{1-4}{4}}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{1-5}{5}}$
$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{-2}{3}}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{-3}{4}}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{-4}{5}}$
$= \,\,\,$ $\dfrac{1}{3} \times 1$ $+$ $\dfrac{1}{4} \times 1$ $+$ $\dfrac{1}{5} \times 1$
$= \,\,\,$ $\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}$
$= \,\,\,$ $\dfrac{20 \times 1 + 15 \times 1 + 12 \times 1}{60}$
$= \,\,\,$ $\dfrac{20+15+12}{60}$
$= \,\,\,$ $\dfrac{47}{60}$
A best free mathematics education website for students, teachers and researchers.
Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.
Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.
Copyright © 2012 - 2022 Math Doubts, All Rights Reserved