
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
162.3k+ 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.
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