
If 0 < p < q , then ${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}}$ is equal to ?
A. e
B. p
C. q
D. 0
Answer
164.1k+ views
Hint: The question is based on limits. First, we must know how to solve the limits. In this question, first, we take common functions, then by subtracting and taking common parts from the powers, we make an equation whose value comes from the 0 < p < q and we get our desired answer.
Complete step by step Solution:
Given that 0 < p < q
We have to find the value of ${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}}$ .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
This means ${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}}$ as p < q$\dfrac{p}{q}$< 1
Now take ${{q}^{n}}$ common from equation (1), we get
${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}} = {{\lim }_{n\to \infty }}{{\left( {{q}^{n}}\left( 1+\dfrac{{{p}^{n}}}{{{q}^{n}}} \right) \right)}^{\dfrac{1}{n}}}$
Now solving the above equation, we get
${{\lim }_{n\to \infty }}{{\left( {{q}^{n}}\left( 1+\dfrac{{{p}^{n}}}{{{q}^{n}}} \right) \right)}^{\dfrac{1}{n}}} = {{e}^{0}}$
Which is equal to q ${{\lim }_{n\to \infty }}{{\left( 1+{{\left( \dfrac{p}{q} \right)}^{n}} \right)}^{\dfrac{1}{n}}}$
Now above equation is equal to q ${{e}^{{{\lim }_{n\to \infty }}\dfrac{1}{n}{{\left( 1+{{\left( \dfrac{p}{q} \right)}^{n}} \right)}^{\dfrac{1}{n}}}}}$
Now we take $\dfrac{1}{n}$ common from the power of the above equation
Then we get q${{e}^{{{\lim }_{n\to \infty }}\dfrac{1}{n}\left( 1+{{\left( \dfrac{p}{q} \right)}^{n}}-1 \right)}}$
That equation is equal to q${{e}^{{{\lim }_{n\to \infty }}\dfrac{1}{n}{{\left( \dfrac{p}{q} \right)}^{n}}}}$
Now we have 0 < p < q i.e. $\dfrac{p}{q}$ < 1
Then we get q${{e}^{{{\lim }_{n\to \infty }}\dfrac{1}{n}{{\left( \dfrac{p}{q} \right)}^{n}}}} = q{{e}^{0}}$
And we know ${{e}^{0}}= 1$
i.e. $q{{e}^{0}} = q$
Hence the value of ${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}} = q$
Hence, the correct option is C.
Note:In this question, first we know to solve the limits. How to open the powers and take powers common also plays a crucial role in solving the question. It is very important to know the values of the functions which have powers. In these types of questions, there are many brackets. We get confused while taking common parts from these brackets. So while solving these types of questions, always take care of solving the powers and the limits.
Complete step by step Solution:
Given that 0 < p < q
We have to find the value of ${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}}$ .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
This means ${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}}$ as p < q$\dfrac{p}{q}$< 1
Now take ${{q}^{n}}$ common from equation (1), we get
${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}} = {{\lim }_{n\to \infty }}{{\left( {{q}^{n}}\left( 1+\dfrac{{{p}^{n}}}{{{q}^{n}}} \right) \right)}^{\dfrac{1}{n}}}$
Now solving the above equation, we get
${{\lim }_{n\to \infty }}{{\left( {{q}^{n}}\left( 1+\dfrac{{{p}^{n}}}{{{q}^{n}}} \right) \right)}^{\dfrac{1}{n}}} = {{e}^{0}}$
Which is equal to q ${{\lim }_{n\to \infty }}{{\left( 1+{{\left( \dfrac{p}{q} \right)}^{n}} \right)}^{\dfrac{1}{n}}}$
Now above equation is equal to q ${{e}^{{{\lim }_{n\to \infty }}\dfrac{1}{n}{{\left( 1+{{\left( \dfrac{p}{q} \right)}^{n}} \right)}^{\dfrac{1}{n}}}}}$
Now we take $\dfrac{1}{n}$ common from the power of the above equation
Then we get q${{e}^{{{\lim }_{n\to \infty }}\dfrac{1}{n}\left( 1+{{\left( \dfrac{p}{q} \right)}^{n}}-1 \right)}}$
That equation is equal to q${{e}^{{{\lim }_{n\to \infty }}\dfrac{1}{n}{{\left( \dfrac{p}{q} \right)}^{n}}}}$
Now we have 0 < p < q i.e. $\dfrac{p}{q}$ < 1
Then we get q${{e}^{{{\lim }_{n\to \infty }}\dfrac{1}{n}{{\left( \dfrac{p}{q} \right)}^{n}}}} = q{{e}^{0}}$
And we know ${{e}^{0}}= 1$
i.e. $q{{e}^{0}} = q$
Hence the value of ${{\lim }_{n\to \infty }}{{({{q}^{n}}+{{p}^{n}})}^{\dfrac{1}{n}}} = q$
Hence, the correct option is C.
Note:In this question, first we know to solve the limits. How to open the powers and take powers common also plays a crucial role in solving the question. It is very important to know the values of the functions which have powers. In these types of questions, there are many brackets. We get confused while taking common parts from these brackets. So while solving these types of questions, always take care of solving the powers and the limits.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
