Examine the continuity of the function

We know that, y = f(x) will be continuous at x = a if,

$\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a^{+}} f(x)$

Given: $f(x)=x^{3}+2 x^{2}-1$

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}(1+h)^{3}+2(1+h)^{2}-1=1+2-1=2$

$\lim _{x \rightarrow 1} f(x)=(1)^{3}+2(1)^{2}-1$

$=1+2-1=2$

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{\rightarrow}(1+h)^{3}+2(1+h)^{2}-1$

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=2$

Hence, $f(x)$ is continuous at $x=1$.