Is the following function continuous at x=2 ?
f\left(x\right)=\begin{cases} x^{2}-3x+5 \ \text{ if }\ x\leqslant2 \cr \cr \dfrac{x+13}{2x+1}\ \text{ if }\ x\gt2 \end{cases}
A function f is continuous in x=a if and only if:
\lim\limits_{x \to a \atop x \lt a}f\left(x\right)=\lim\limits_{x \to a \atop x \gt a}f\left(x\right)=f\left(a\right)
In our problem:
\lim\limits_{x \to 2 \atop x \lt 2}f\left(x\right)=\lim\limits_{x \to 2 \atop x \lt 2}\left( x^{2}-3x+5 \right)=\left(2\right)^{2}-3\left(2\right)+5=3
\lim\limits_{x \to 2 \atop x \gt 2}=\lim\limits_{x \to 2 \atop x \gt 2}\left( \dfrac{x+13}{2x+1} \right)=\dfrac{\left(2\right)+13}{2\left(2\right)+1}=3
f\left(2\right)=\left(2\right)^{2}-3\left(2\right)+5=3
Therefore:
\lim\limits_{x \to 2 \atop x \lt 2}f\left(x\right)=\lim\limits_{x \to 2 \atop x \gt 2}f\left(x\right)=f\left(2\right)
We conclude that f is continuous at x=2.
Is the following function continuous at x=3 ?
f\left(x\right)=\begin{cases} \sqrt{x^{2}+5x+12} \ \text{ if }\ x\leqslant3 \cr \cr 7x-15\ \text{ if }\ x\gt3 \end{cases}
A function f is continuous in x=a if and only if:
\lim\limits_{x \to a \atop x \lt a}f\left(x\right)=\lim\limits_{x \to a \atop x \gt a}f\left(x\right)=f\left(a\right)
In our problem:
\lim\limits_{x \to 3 \atop x \lt 3}f\left(x\right)=\lim\limits_{x \to 3 \atop x \lt 3}\left( \sqrt{x^{2}+5x+12}\right)=\sqrt{\left(3\right)^{2}+5\left(3\right)+12}=6
\lim\limits_{x \to 3 \atop x \gt 3}=\lim\limits_{x \to 3 \atop x \gt 3}\left( 7x-15 \right)=7\left(3\right)-15=6
f\left(3\right)=\sqrt{\left(3\right)^{2}+5\left(3\right)+12}=6
Therefore:
\lim\limits_{x \to 3 \atop x \lt 3}f\left(x\right)=\lim\limits_{x \to 3 \atop x \gt 3}f\left(x\right)=f\left(3\right)
We conclude that f is continuous at x=3.
Is the following function continuous at x=2 ?
f\left(x\right)=\begin{cases} \dfrac{\sin\left(x-2\right)}{x-2}\ \text{ if }\ x\lt2 \cr \cr \log_3\left(x^{2}-1\right)\ \text{ if }\ x\geqslant2 \end{cases}
A function f is continuous in x=a if and only if:
\lim\limits_{x \to a \atop x \lt a}f\left(x\right)=\lim\limits_{x \to a \atop x \gt a}f\left(x\right)=f\left(a\right)
In our problem:
\lim\limits_{x \to 2 \atop x \lt 2}f\left(x\right)=\lim\limits_{x \to 2 \atop x \lt 2}\left( \dfrac{\sin\left(x-2\right)}{x-2}\right)=1
\lim\limits_{x \to 2 \atop x \gt 2}=\lim\limits_{x \to 2 \atop x \gt 2}\left( \log_3\left(x^{2}-1\right)\right)=\log_3\left(3\right)=1
f\left(2\right)=\log_3\left(3\right)=1
Therefore:
\lim\limits_{x \to 2 \atop x \lt 2}f\left(x\right)=\lim\limits_{x \to 2 \atop x \gt 2}f\left(x\right)=f\left(2\right)
We conclude that f is continuous at x=2.
Is the following function continuous at x=-1 ?
f\left(x\right)=\begin{cases} \dfrac{3x-5}{x-3}\ \text{ if }\ x\lt-1 \cr \cr x+3 \text{ if }\ x\geqslant-1 \end{cases}
A function f is continuous in x=a if and only if:
\lim\limits_{x \to a \atop x \lt a}f\left(x\right)=\lim\limits_{x \to a \atop x \gt a}f\left(x\right)=f\left(a\right)
In our problem:
\lim\limits_{x \to -1 \atop x \lt -1}f\left(x\right)=\lim\limits_{x \to -1 \atop x \lt -1}\left( \dfrac{3x-5}{x-3}\right)=2
\lim\limits_{x \to -1 \atop x \gt -1}=\lim\limits_{x \to -1 \atop x \gt -1}\left( x+3\right)=2
f\left(-1\right)=\left(-1\right)+3=2
Therefore:
\lim\limits_{x \to -1 \atop x \lt -1}f\left(x\right)=\lim\limits_{x \to -1 \atop x \gt -1}f\left(x\right)=f\left(-1\right)
We conclude that f is continuous at x=-1.
Is the following function continuous at x=\dfrac{7\pi}{9} ?
f\left(x\right)=\begin{cases}\cos\left(3x-2\pi\right)\ \text{ if }\ x\lt\dfrac{7\pi}{9} \cr \cr \sin\left(\dfrac{9x}{7}\right)+\dfrac{1}{2}\text{ if }\ x\geqslant\dfrac{7\pi}{9} \end{cases}
A function f is continuous in x=a if and only if:
\lim\limits_{x \to a \atop x \lt a}f\left(x\right)=\lim\limits_{x \to a \atop x \gt a}f\left(x\right)=f\left(a\right)
In our problem:
\lim\limits_{x \to \frac{7\pi}{9} \atop x \lt \frac{7\pi}{9} }f\left(x\right)=\lim\limits_{x \to \frac{7\pi}{9} \atop x \lt \frac{7\pi}{9} }\left( \cos\left(3x-2\pi\right)\right)=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}
\lim\limits_{x \to \frac{7\pi}{9} \atop x \gt \frac{7\pi}{9} }=\lim\limits_{x \to \frac{7\pi}{9} \atop x \gt \frac{7\pi}{9} }\left( \sin\left(\dfrac{9x}{7}\right)+\dfrac{1}{2}\right)=\sin\left(\pi\right)+\dfrac{1}{2}=\dfrac{1}{2}
f\left( \dfrac{7\pi}{9} \right)=\sin\left(\pi\right)+\dfrac{1}{2}=\dfrac{1}{2}
Therefore:
\lim\limits_{x \to \frac{7\pi}{9} \atop x \lt \frac{7\pi}{9}}f\left(x\right)=\lim\limits_{x \to \frac{7\pi}{9} \atop x \gt \frac{7\pi}{9}}f\left(x\right)=f\left( \dfrac{7\pi}{9} \right)
We conclude that f is continuous at x=\dfrac{7\pi}{9}.
Is the following function continuous at x=4 ?
f\left(x\right)=\begin{cases}\sqrt{2x+1} \text{ if }\ x\lt4\cr \cr \dfrac{5x+4}{x+2}\text{ if }\ x\geqslant4 \end{cases}
A function f is continuous in x=a if and only if:
\lim\limits_{x \to a \atop x \lt a}f\left(x\right)=\lim\limits_{x \to a \atop x \gt a}f\left(x\right)=f\left(a\right)
In our problem:
\lim\limits_{x \to 4\atop x \lt 4 }f\left(x\right)=\lim\limits_{x \to 4 \atop x \lt 4}\sqrt{2x+1}=\sqrt{9}=3
\lim\limits_{x \to 4 \atop x \gt 4 }=\lim\limits_{x \to 4 \atop x \gt4 }\left( \dfrac{5x+4}{x+2}\right)=4
f\left( 4 \right)=\dfrac{5\left(4\right)+4}{\left(4\right)+2}=4
Therefore:
\lim\limits_{x \to4\atop x \lt 4}f\left(x\right)\neq\lim\limits_{x \to 4 \atop x \gt 4}f\left(x\right)
We conclude that f is not continuous at x=4.
Is the following function continuous at x=1 ?
f\left(x\right)=\begin{cases}\dfrac{2x^{2}+3x-5}{x-6} \text{ if }\ x\lt1\cr \cr \ln\left(x\right)+3\text{ if }\ x\geqslant1 \end{cases}
A function f is continuous in x=a if and only if:
\lim\limits_{x \to a \atop x \lt a}f\left(x\right)=\lim\limits_{x \to a \atop x \gt a}f\left(x\right)=f\left(a\right)
In our problem:
\lim\limits_{x \to 1\atop x \lt 1 }f\left(x\right)=\lim\limits_{x \to 1 \atop x \lt 1}\dfrac{2x^{2}+3x-5}{x-6}=0
\lim\limits_{x \to 1 \atop x \gt 1 }=\lim\limits_{x \to 1 \atop x \gt1 }\left( \ln\left(x\right)+3\right)=3
f\left( 1 \right)=\ln\left(1\right)+3=3
Therefore:
\lim\limits_{x \to1\atop x \lt 1}f\left(x\right)\neq\lim\limits_{x \to 1 \atop x \gt 1}f\left(x\right)
We conclude that f is not continuous at x=1.