Evaluate the limit, if it exists. If a limit does not exist, type "DNE".

[tex]\[
\lim_{y \rightarrow 0}\left(\frac{2}{y^2+y}-\frac{2}{y}\right)
\][/tex]

Limit: [tex]$\square$[/tex]



Answer :

To evaluate the limit [tex]\(\lim _{y \rightarrow 0}\left(\frac{2}{y^2+y}-\frac{2}{y}\right)\)[/tex], let's break it down step by step.

Given the expression:
[tex]\[ \frac{2}{y^2 + y} - \frac{2}{y} \][/tex]

Let's combine the fractions into a single expression:
[tex]\[ \frac{2}{y^2 + y} - \frac{2}{y} = \frac{2y - 2(y^2 + y)}{y(y^2 + y)} \][/tex]

First, find a common denominator, which is [tex]\(y(y^2 + y)\)[/tex]:
[tex]\[ \frac{2}{y^2 + y} - \frac{2}{y} = \frac{2y - 2(y^2 + y)}{y(y^2 + y)} \][/tex]

Now, simplify the numerator:
[tex]\[ 2y - 2(y^2 + y) = 2y - 2y^2 - 2y = -2y^2 \][/tex]

Thus, the expression becomes:
[tex]\[ \frac{-2y^2}{y(y^2 + y)} = \frac{-2y^2}{y^3 + y^2} \][/tex]

Further simplifying the numerator and denominator:
[tex]\[ \frac{-2y^2}{y^3 + y^2} = \frac{-2y^2}{y^2(y + 1)} = \frac{-2y}{y + 1} \][/tex]

Now, we need to evaluate the limit of this simplified expression as [tex]\( y \)[/tex] approaches 0:
[tex]\[ \lim_{y \to 0} \frac{-2y}{y + 1} \][/tex]

Substitute [tex]\( y = 0 \)[/tex]:
[tex]\[ \frac{-2 \cdot 0}{0 + 1} = 0 \][/tex]

Therefore, the limit is:
[tex]\[ \boxed{-2} \][/tex]