Math Doubts

Evaluate $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{\sqrt[3]{x}+\sqrt[4]{x}+\sqrt[5]{x}-3}{x-1}}$

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.

Write the value of Radical symbol

$\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{\sqrt[3]{x}+\sqrt[4]{x}+\sqrt[5]{x}-3}{x-1}}$

In numerator, the first three terms have different radical symbols and they can be expressed in exponential form.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}+x^\frac{1}{4}+x^\frac{1}{5}-3}{x-1}}$

Convert the function same as Limit Rule

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.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}-1}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{4}-1}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{5}-1}{x-1}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}-{(1)}^\frac{1}{3}}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{4}-{(1)}^\frac{1}{4}}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{5}-{(1)}^\frac{1}{5}}{x-1}}$

Evaluate the Limit of Each function

$\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{3}-{(1)}^\frac{1}{3}}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{4}-{(1)}^\frac{1}{4}}{x-1}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^\frac{1}{5}-{(1)}^\frac{1}{5}}{x-1}}$

Now, evaluate limit of each function as per limit of (xn-an)/(x-a) as x approaches a rule.

$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{1}{3}-1}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{1}{4}-1}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{1}{5}-1}$

$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{1-3}{3}}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{1-4}{4}}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{1-5}{5}}$

$= \,\,\,$ $\dfrac{1}{3} \times {(1)}^{\frac{-2}{3}}$ $+$ $\dfrac{1}{4} \times {(1)}^{\frac{-3}{4}}$ $+$ $\dfrac{1}{5} \times {(1)}^{\frac{-4}{5}}$

$= \,\,\,$ $\dfrac{1}{3} \times 1$ $+$ $\dfrac{1}{4} \times 1$ $+$ $\dfrac{1}{5} \times 1$

$= \,\,\,$ $\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}$

$= \,\,\,$ $\dfrac{20 \times 1 + 15 \times 1 + 12 \times 1}{60}$

$= \,\,\,$ $\dfrac{20+15+12}{60}$

$= \,\,\,$ $\dfrac{47}{60}$

Ashok Kumar B.E. - Founder of Math Doubts

Ashok Kumar, B.E.

Founder of Math Doubts

A Specialist in Mathematics, Physics, and Engineering, dedicated to helping students master math concepts from basics to advanced levels with clarity and precision.