Answer
Verified
382.8k+ views
Hint: In math, a limit can be defined such a way that the function or series tends as the input tends to about some unique value. Limits are very important in the use of calculus and analysis of mathematical modeling, also it is used from different kinds of defined derivatives, continuity, and integrals. Here we have to solve the limit for the greatest integer function.
Complete step-by-step solution:
Step 1:
First we write the given data
\[\mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right]\]
Now we know that
\[x \geqslant \left[ x \right] \geqslant x - 1\,\,\,\forall \,\,\,x \in R\]
Then we put
\[\Rightarrow x = \dfrac{1}{{{x^3}}}\]
\[\Rightarrow x \geqslant \left[ x \right] \geqslant x - 1\,\,\,\forall \,\,\,x \in R\]
\[\Rightarrow \dfrac{1}{{{x^3}}} \geqslant \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \dfrac{1}{{{x^3}}} - 1\,\,\,\forall \,\,\,x \in R\]
Then multiply it by \[{x^8}\] so
\[{x^8} \times \dfrac{1}{{{x^3}}} \geqslant {x^8} \times \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant {x^8} \times \left( {\dfrac{1}{{{x^3}}} - 1} \right)\,\,\,\forall \,\,\,but\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} {x^{8 - 3}} = \mathop {\lim }\limits_{x \to 0} {x^5}\]
Then put the limit
\[ = \mathop {\lim }\limits_{x \to 0} {x^5} = {0^5} = 0\]
So take the limits below
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} {x^8}\dfrac{1}{{{x^3}}} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8} \times \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{x^8}}}{{{x^3}}} - {x^8}} \right)\,\,\,\forall \,\,x \in R\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} - \mathop {\lim }\limits_{x \to 0} {x^8}\,\,\,\forall \,\,x \in R\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} {x^5} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} {x^5} - \mathop {\lim }\limits_{x \to 0} {x^8}\,\,\,\forall \,\,x \in R\]
Now put the limits then we got
\[\Rightarrow {0^5} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant {0^5} - {0^8}\,\,\,\forall \,\,x \in R\]
\[\Rightarrow 0 \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant 0 - 0\,\,\,\forall \,\,x \in R\]
\[\Rightarrow 0 \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant 0\,\,\,\forall \,\,x \in R\]
Or we can write it as
\[\mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] = 0\]
So the correct option are
b) A rational number
c) An integer
d) Zero
Note: The given options are correct for the following reasons, option (b) is correct due to the A Rational number due to the ratio of the power are different and put as the integer in the numerator as well as a denominator so the option is correct and then the option (c) is correct due to the in the numerator and the denominator the having the power when we put the integer than integer power is also integer hence the option (c) is correct and the last and final option (d) is correct due to our solution that provided above lead the limiting value of the given question is zero hence the option (d) is correct.
Complete step-by-step solution:
Step 1:
First we write the given data
\[\mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right]\]
Now we know that
\[x \geqslant \left[ x \right] \geqslant x - 1\,\,\,\forall \,\,\,x \in R\]
Then we put
\[\Rightarrow x = \dfrac{1}{{{x^3}}}\]
\[\Rightarrow x \geqslant \left[ x \right] \geqslant x - 1\,\,\,\forall \,\,\,x \in R\]
\[\Rightarrow \dfrac{1}{{{x^3}}} \geqslant \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \dfrac{1}{{{x^3}}} - 1\,\,\,\forall \,\,\,x \in R\]
Then multiply it by \[{x^8}\] so
\[{x^8} \times \dfrac{1}{{{x^3}}} \geqslant {x^8} \times \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant {x^8} \times \left( {\dfrac{1}{{{x^3}}} - 1} \right)\,\,\,\forall \,\,\,but\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} {x^{8 - 3}} = \mathop {\lim }\limits_{x \to 0} {x^5}\]
Then put the limit
\[ = \mathop {\lim }\limits_{x \to 0} {x^5} = {0^5} = 0\]
So take the limits below
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} {x^8}\dfrac{1}{{{x^3}}} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8} \times \left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{x^8}}}{{{x^3}}} - {x^8}} \right)\,\,\,\forall \,\,x \in R\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^8}}}{{{x^3}}} - \mathop {\lim }\limits_{x \to 0} {x^8}\,\,\,\forall \,\,x \in R\]
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} {x^5} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant \mathop {\lim }\limits_{x \to 0} {x^5} - \mathop {\lim }\limits_{x \to 0} {x^8}\,\,\,\forall \,\,x \in R\]
Now put the limits then we got
\[\Rightarrow {0^5} \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant {0^5} - {0^8}\,\,\,\forall \,\,x \in R\]
\[\Rightarrow 0 \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant 0 - 0\,\,\,\forall \,\,x \in R\]
\[\Rightarrow 0 \geqslant \mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] \geqslant 0\,\,\,\forall \,\,x \in R\]
Or we can write it as
\[\mathop {\lim }\limits_{x \to 0} {x^8}\left[ {\dfrac{1}{{{x^3}}}} \right] = 0\]
So the correct option are
b) A rational number
c) An integer
d) Zero
Note: The given options are correct for the following reasons, option (b) is correct due to the A Rational number due to the ratio of the power are different and put as the integer in the numerator as well as a denominator so the option is correct and then the option (c) is correct due to the in the numerator and the denominator the having the power when we put the integer than integer power is also integer hence the option (c) is correct and the last and final option (d) is correct due to our solution that provided above lead the limiting value of the given question is zero hence the option (d) is correct.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Draw a labelled sketch of the human eye class 12 physics CBSE