Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Find the  \lim_{x\rightarrow a}\left ( HM \right )  where HM is the harmonic mean of the positive n real numbers x_{1},x_{2},x_{3},\cdots x_{n}, and \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\cdots +\frac{1}{x_{n}}= \lim _{x \rightarrow a}f\left ( x \right )= K^{n}.

Option: 1

K^{n}


Option: 2

\frac{1}{K^{n}}


Option: 3

\frac{n}{K^{n}}


Option: 4

\left ( \frac{n}{K} \right )^{n}


Answers (1)

best_answer

Note that the “Constant multiple law for limits” indicates that \lim _{x \rightarrow a}cf\left ( x \right )= c\cdot \lim _{x \rightarrow a}f\left ( x \right )

The harmonic mean of the positive n real numbers x_{1},x_{2},x_{3},\dots x_{n} is 

HM= \frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\dots+\frac{1}{x_{n}}}\quad \dots\left ( i \right )

Again, the provided limit is
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\dots+\frac{1}{x_{n}}= \lim _{x \rightarrow a}f\left ( x \right )= K^{n}\quad \dots\left ( ii \right )

Derive the following limit using the equation (i), (ii) and applying the Constant multiple law for limits.

\lim _{x \rightarrow a}\left ( HM \right )
= \lim _{x \rightarrow a}\left [ \frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\dots+\frac{1}{x_{n}}} \right ]
=\frac{\lim _{x \rightarrow a}n}{\lim _{x \rightarrow a}\left [ \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\dots+\frac{1}{x_{n}}\right ]}
=\frac{n}{K^{n}}

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 chemistry needs conceptual understanding; 5 tips to score high
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon; paper pattern, exams from January 22
anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters

Latest Question