Question

How do you evaluate the limit of $\mathop {\lim }\limits_{n \to 0} \dfrac{{2{n^2}}}{n}$? ${a^2} = a \times a$.

Answer 1

This question can be solved by using fact and applying the identity inside the given limit, and then we have to simplify further and then we can make use of the fact $\mathop {\lim }\limits_{x \to a} \dfrac{1}{{f\left( x \right) + 1}} = \dfrac{1}{{f\left( a \right) + 1}}$ and calculate the value and we will get the required answer.

Complete step by step solution:
A limit is defined as a function that has some value that approaches the input. A limit of a function is represented as :
$\mathop {\lim }\limits_{x \to n} f\left( x \right) = L$,
Here lim refers to limit, it generally describes that the real valued function $f\left( x \right)$ tends to attain the limit L as $x$ tends to n and is denoted by an arrow.
We can read this as “the limit of any given function ‘f’ of ‘$x$’ as ‘$x$’ approaches to$n$ is equal to L.
Given expression is $\mathop {\lim }\limits_{n \to 0} \dfrac{{2{n^2}}}{n}$,
Applying limit we get,
$\mathop {\lim }\limits_{n \to 0} \dfrac{{2{n^2}}}{n}$,
Now applying the fact ${a^2} = a \times a$ to the numerator we get,
$\Rightarrow \mathop {\lim }\limits_{n \to 0} \dfrac{{2 \times n \times n}}{n}$,
Now eliminating the like terms we get,
$\Rightarrow \mathop {\lim }\limits_{n \to 0} 2n$,
Now using the fact $\mathop {\lim }\limits_{x \to a} \dfrac{1}{{f\left( x \right) + 1}} = \dfrac{1}{{f\left( a \right) + 1}}$,
$\Rightarrow 2\left( 0 \right)$,
Now simplifying we get,
$\Rightarrow \mathop {\lim }\limits_{n \to 0} \dfrac{{2{n^2}}}{n} = 0$,
So the limit value for $\mathop {\lim }\limits_{n \to 0} \dfrac{{2{n^2}}}{n}$is 0.

Final Answer:
$\therefore$The value of the limit$\mathop {\lim }\limits_{n \to 0} \dfrac{{2{n^2}}}{n}$ is 0.

Note:
We should not directly substitute the value of 0 in the given limit as we can see that it results in an indeterminate form $\infty \to \infty$. We should not assume $\left[ . \right]$ as a greater 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 $\infty$.