Continuity

${\lim }_{x\to 0^{-}}(x+1)=1$, ${\lim }_{x\to 0^{+}}(x^2+1)=1$, $f(0)=1$. All three agree, so $f$ is continuous at $0$.

Polynomial functions are continuous everywhere on $\mathbb{R}$.

$f$ is undefined at $x=2$, where the denominator vanishes.

(a) A polynomial — continuous on all of $\mathbb{R}$ ✓.

(b) Discontinuous at $x=0$ (not defined).

(c) Discontinuous at $x=\frac{\pi }{2}+k\pi$.

(d) Defined only for $x\ge 0$, so not continuous on negative reals.

Left limit: ${\displaystyle \underset{x\to 2^{-}}{\lim }(x+1)=3}$.

Right limit: ${\displaystyle \underset{x\to 2^{+}}{\lim }(x^2-1)=3}$.

$f(2)=2^2-1=3$.

All three agree, so $f$ is continuous at $x=2$.

Continuity is preserved by sum, difference, product, and composition (provided $g$ is continuous at $x_0$ and $f$ is continuous at $g(x_0)$).

${\lim }_{x\to 0^{-}}f=-1$ and ${\lim }_{x\to 0^{+}}f=1$ — finite but unequal limits, hence a jump discontinuity.

LHS: $2\cdot 1+1=3$. RHS: $1+a$. Setting equal: $a=2$.

${\displaystyle \frac{x^2-1}{x-1}}=x+1\to 2$ as $x\to 1$. For continuity, $a=2$.

$f(0)=-1<0$ and $f(1)=1>0$. Since $f$ is continuous, by the IVT there exists $c\in (0,1)$ with $f(c)=0$.