To check if the limit exists, let's try inputting x = -2 (the lmiit) into the expression and see if it is defined:
[tex]\begin{gathered} x=-2 \\ \frac{2+\mleft(-2\mright)}{(-2)^{2+2(-2)}}=\frac{2-2}{(-2)^{2-4}}=\frac{0}{(-2)^{-2}}=0\cdot(-2)^2=0\cdot4=0 \end{gathered}[/tex]
There is no problem with this reasoning, for example, there is no zero in the denominator or squaare root of a negative number, so the limit is the same as simply substituting x = -2.
So the answer is:
[tex]\lim _{x\to-2}\frac{2+x}{x^{2+2x}}=\frac{2+(-2)}{(-2)^{2+2(-2)}}=0[/tex]
