How do you evaluate the function $ \left[ \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{{{x}^{2}}+x} \right) \right] $ as x approaches to zero?
Answer
630.3k+ views
Hint: We start solving the problem by performing the subtraction operation to the functions present inside the given limit. We then make the necessary calculations and then take the cancel the common factors present in both numerator and denominator present inside the limit. We then make use of the fact that $ \displaystyle \lim_{x \to a}\dfrac{1}{f\left( x \right)+1}=\dfrac{1}{f\left( a \right)+1} $ and then make the necessary calculations to get the required answer.
Complete step by step answer:
According to the problem, we are asked to find the value of the given function $ \left[ \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{{{x}^{2}}+x} \right) \right] $ as x approaches zero.
Let us assume $ L=\displaystyle \lim_{x \to 0}\left[ \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{{{x}^{2}}+x} \right) \right] $ .
Let us subtract the given functions inside the limit.
$ \Rightarrow L=\displaystyle \lim_{x \to 0}\left[ \dfrac{{{x}^{2}}+x-x}{x\left( {{x}^{2}}+x \right)} \right] $ .
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\left[ \dfrac{{{x}^{2}}+x-x}{{{x}^{2}}\left( x+1 \right)} \right]\].
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\left[ \dfrac{{{x}^{2}}}{{{x}^{2}}\left( x+1 \right)} \right]\].
We can see that the numerator and denominator inside the limit have a common factor $ {{x}^{2}} $.
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\left[ \dfrac{1}{x+1} \right]\].
We know that $ \displaystyle \lim_{x \to a}\dfrac{1}{f\left( x \right)+1}=\dfrac{1}{f\left( a \right)+1} $ .
\[\Rightarrow L=\left[ \dfrac{1}{0+1} \right]\].
\[\Rightarrow L=\left[ \dfrac{1}{1} \right]\].
\[\Rightarrow L=1\].
We have found the value of the given function when the value of x approaches zero as 1.
$ \, therefore, $ The value of the given function when the value of x approaches zero is 1.
Note:
We should not directly substitute the value 0 in the given limit as we can see that it results to be an indeterminant form $ \infty -\infty $. We should not assume $ \left[ . \right] $ as a Greatest Integer Function unless it is mentioned in the problem. Whenever we get this type of problem, we first simplify the function given inside the limit and substitute the value of the given to get the required answer. Similarly, we can expect problems to find the limit of the given function when x approaches $ \infty $.
Complete step by step answer:
According to the problem, we are asked to find the value of the given function $ \left[ \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{{{x}^{2}}+x} \right) \right] $ as x approaches zero.
Let us assume $ L=\displaystyle \lim_{x \to 0}\left[ \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{{{x}^{2}}+x} \right) \right] $ .
Let us subtract the given functions inside the limit.
$ \Rightarrow L=\displaystyle \lim_{x \to 0}\left[ \dfrac{{{x}^{2}}+x-x}{x\left( {{x}^{2}}+x \right)} \right] $ .
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\left[ \dfrac{{{x}^{2}}+x-x}{{{x}^{2}}\left( x+1 \right)} \right]\].
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\left[ \dfrac{{{x}^{2}}}{{{x}^{2}}\left( x+1 \right)} \right]\].
We can see that the numerator and denominator inside the limit have a common factor $ {{x}^{2}} $.
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\left[ \dfrac{1}{x+1} \right]\].
We know that $ \displaystyle \lim_{x \to a}\dfrac{1}{f\left( x \right)+1}=\dfrac{1}{f\left( a \right)+1} $ .
\[\Rightarrow L=\left[ \dfrac{1}{0+1} \right]\].
\[\Rightarrow L=\left[ \dfrac{1}{1} \right]\].
\[\Rightarrow L=1\].
We have found the value of the given function when the value of x approaches zero as 1.
$ \, therefore, $ The value of the given function when the value of x approaches zero is 1.
Note:
We should not directly substitute the value 0 in the given limit as we can see that it results to be an indeterminant form $ \infty -\infty $. We should not assume $ \left[ . \right] $ as a Greatest Integer Function unless it is mentioned in the problem. Whenever we get this type of problem, we first simplify the function given inside the limit and substitute the value of the given to get the required answer. Similarly, we can expect problems to find the limit of the given function when x approaches $ \infty $.
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