
How do you find the limit of \[\dfrac{\dfrac{1}{y}-\dfrac{1}{7}}{y-7}\] as y approaches 7?
Answer
531.6k+ views
Hint: In the given question, we have asked to simplify an expression and the limit of that expression is given. In order to solve the question, first we need to simplify the numerator and denominator one by one of the given expressions. Later after simplifying we will cancel out the common factor. Then we put y = 7 in the simplified expression. In this way we will get the answer to this question.
Complete step by step solution:
We have given that,
\[\Rightarrow \dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}\]
Simplifying the numerator;
We have the numerator as follows,
\[\left( \dfrac{1}{y}-\dfrac{1}{7} \right)\]
Taking the LCM of the above equation, we get
\[\left( \dfrac{7-y}{7y} \right)\]
Now,
Combining the numerator and denominator,
We have,
\[\Rightarrow \dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=\dfrac{\left( \dfrac{7-y}{7y} \right)}{y-7}=\dfrac{7-y}{7y}\times \dfrac{1}{y-7}\]
Rewrite the simplified expression as, we get
\[\Rightarrow \dfrac{7-y}{7y}\times \dfrac{1}{y-7}=\dfrac{-1\left( y-7 \right)}{7y}\times \dfrac{1}{y-7}\]
Cancelling out the common factor of the above expression, we get
\[\Rightarrow -\dfrac{1}{7y}\]
Now, putting the limit;
\[\underset{y\to 7}{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=\underset{y\to 7}{\mathop{\lim }}\,-\dfrac{1}{7y}\]
Putting the value of y equals to 7 in the above expression, we get
\[\underset{y\to 7}{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=\underset{y\to 7}{\mathop{\lim }}\,-\dfrac{1}{7y}=-\dfrac{1}{7\times 7}\]
Solving the numbers in the above expression, we get
\[\underset{y\to 7}{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=\underset{y\to 7}{\mathop{\lim }}\,-\dfrac{1}{7y}=-\dfrac{1}{7\times 7}=-\dfrac{1}{49}\]
Therefore, \[\underset{y\to 7}{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=-\dfrac{1}{49}\] Hence, this is the required answer.
Note: While solving these types of problems, students need to be very careful while doing the calculation part to avoid making any type of error. They need to know about the concept of the simplification of the quadratic equation. Instead of factorizing the polynomial we can solve this question by derivate the numerator and denominator with respect to ‘y’ and then put the given value of ‘y’, this is known as l’hospital rule. Both the ways we will get the same answer.
Complete step by step solution:
We have given that,
\[\Rightarrow \dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}\]
Simplifying the numerator;
We have the numerator as follows,
\[\left( \dfrac{1}{y}-\dfrac{1}{7} \right)\]
Taking the LCM of the above equation, we get
\[\left( \dfrac{7-y}{7y} \right)\]
Now,
Combining the numerator and denominator,
We have,
\[\Rightarrow \dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=\dfrac{\left( \dfrac{7-y}{7y} \right)}{y-7}=\dfrac{7-y}{7y}\times \dfrac{1}{y-7}\]
Rewrite the simplified expression as, we get
\[\Rightarrow \dfrac{7-y}{7y}\times \dfrac{1}{y-7}=\dfrac{-1\left( y-7 \right)}{7y}\times \dfrac{1}{y-7}\]
Cancelling out the common factor of the above expression, we get
\[\Rightarrow -\dfrac{1}{7y}\]
Now, putting the limit;
\[\underset{y\to 7}{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=\underset{y\to 7}{\mathop{\lim }}\,-\dfrac{1}{7y}\]
Putting the value of y equals to 7 in the above expression, we get
\[\underset{y\to 7}{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=\underset{y\to 7}{\mathop{\lim }}\,-\dfrac{1}{7y}=-\dfrac{1}{7\times 7}\]
Solving the numbers in the above expression, we get
\[\underset{y\to 7}{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=\underset{y\to 7}{\mathop{\lim }}\,-\dfrac{1}{7y}=-\dfrac{1}{7\times 7}=-\dfrac{1}{49}\]
Therefore, \[\underset{y\to 7}{\mathop{\lim }}\,\dfrac{\left( \dfrac{1}{y}-\dfrac{1}{7} \right)}{y-7}=-\dfrac{1}{49}\] Hence, this is the required answer.
Note: While solving these types of problems, students need to be very careful while doing the calculation part to avoid making any type of error. They need to know about the concept of the simplification of the quadratic equation. Instead of factorizing the polynomial we can solve this question by derivate the numerator and denominator with respect to ‘y’ and then put the given value of ‘y’, this is known as l’hospital rule. Both the ways we will get the same answer.
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