Limit and Continuity

A function $f(x)$ is continuous at $x=a$ if

$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a).$

That is, the left-hand limit, the right-hand limit and the value of the function at $x=a$ all exist and are equal.

Condition for continuity at $x = 2$: A function $f(x)$ is continuous at $x = a$ if

$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

Left-hand limit (LHL) at $x = 2$:

$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 2) = 2 + 2 = 4$

Right-hand limit (RHL) at $x = 2$:

$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 2) = 3(2) - 2 = 4$

Value of the function at $x = 2$:

$f(2) = 4$

Conclusion:

$\text{LHL} = \text{RHL} = f(2) = 4$

Since the left-hand limit, the right-hand limit and the value of the function at $x = 2$ are all equal to $4$, the function $f(x)$ is continuous at $x = 2$.

(a)
$f(x)=3x+2$

$f(2.999)=3(2.999)+2=8.997+2=10.997$

$f(3.001)=3(3.001)+2=9.003+2=11.003$

So,
$f(2.999)=10.997,\qquad f(3.001)=11.003$

(b)
Yes, the function $f(x)=3x+2$ is continuous at $x=3$.

Reason:
$f(3)=3(3)+2=11$

Also, as $x$ approaches 3 from both sides, the values of $f(x)$ approach 11.
Here,
$f(2.999)=10.997 \quad \text{and} \quad f(3.001)=11.003$

Both are very close to $f(3)=11$. Hence,
$\lim_{x\to 3^-} f(x)=\lim_{x\to 3^+} f(x)=f(3)=11$

Therefore, $f(x)$ is continuous at $x=3$.

Given: $f(x) = \begin{cases} 3x - 1 & \text{for } x \ge 2 \\ x + 3 & \text{for } x < 2 \end{cases}$

To check: continuity at $x = 2$.

Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2} (x + 3) = 2 + 3 = 5$

Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2} (3x - 1) = 3(2) - 1 = 5$

Value of the function at $x = 2$: $f(2) = 3(2) - 1 = 5 \quad (\text{since } x = 2 \text{ falls in } x \ge 2)$

Conclusion: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 5$

Since the left-hand limit, the right-hand limit and the value of the function at $x = 2$ are all equal, the function $f(x)$ is continuous at $x = 2$.

For $f(x)=3x+5$:

$f(1.99)=3(1.99)+5=10.97$ $f(2.01)=3(2.01)+5=11.03$ $f(2)=3(2)+5=11$

As $x$ approaches $2$ from the left (1.99) and from the right (2.01), the values approach $11$. So $\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=11=f(2)$.

Logic: the function is defined at $x=2$, the limit exists, and the limit equals the function value. Hence $f(x)$ is continuous at $x=2$.

A function $f(x)$ is continuous at $x=a$ if

$\lim_{x\to a^-} f(x)=\lim_{x\to a^+} f(x)=f(a)$

that is, $\lim_{x\to a} f(x)=f(a)$.

The set of real numbers ($\mathbb{R}$) is continuous.

$\lim_{x\to a^-} f(x)$ is read as: "the limit of $f(x)$ as $x$ approaches $a$ from the left (from values less than $a$)", i.e. the value that $f(x)$ gets closer and closer to as $x$ approaches $a$ through values smaller than $a$. It is called the left-hand limit (LHL).

Given:

$f(x) = \begin{cases} 2x + 4 & \text{for } x < 3 \\ 4x - 2 & \text{for } x \ge 3 \end{cases}$

Left-hand limit at $x = 3$:

$\lim_{x \to 3^-} f(x) = 2(3) + 4 = 10$

Right-hand limit at $x = 3$:

$\lim_{x \to 3^+} f(x) = 4(3) - 2 = 10$

Value of the function at $x = 3$:

$f(3) = 4(3) - 2 = 10 \quad (\text{since } x = 3 \text{ falls in } x \ge 3)$

Conclusion:

$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) = 10$

Since the left-hand limit, the right-hand limit and the value of the function at $x = 3$ are all equal, the function $f(x)$ is continuous at $x = 3$.

(a)
$f(1.999)=2(1.999)+3=3.998+3=6.998$

(b)
$f(2.001)=2(2.001)+3=4.002+3=7.002$

(c)
Yes, (f(x)=2x+3) is continuous at (x=2).

Reason: $f(2)=2(2)+3=7$ Also, $\lim_{x\to 2^-} f(x)=7,\qquad \lim_{x\to 2^+} f(x)=7$ So, $\lim_{x\to 2} f(x)=f(2)=7$ Hence, the function is continuous at (x=2).

Given $f(x)=4x+1$.

(i) $f(1.999)=4(1.999)+1=8.996$ and $f(2.001)=4(2.001)+1=9.004$.

(ii) As $x$ approaches 2 from either side, $f(x)$ approaches 9. So

$\lim_{x\to 2^-}f(x)=9,\qquad \lim_{x\to 2^+}f(x)=9$

(iii) $f(2)=4(2)+1=9$.

(iv) Yes. Here

$\lim_{x\to 2^-}f(x)=\lim_{x\to 2^+}f(x)=f(2)=9$

The left limit, right limit and the value of the function at $x=2$ are all equal, so $f(x)$ is continuous at $x=2$.

Given $f(x)=3x+2$ at $x=3$.

Left hand limit: $\lim_{x\to3^-}(3x+2)=3(3)+2=11$

Right hand limit: $\lim_{x\to3^+}(3x+2)=3(3)+2=11$

Functional value: $f(3)=3(3)+2=11$

Since $\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=f(3)=11$, the function $f(x)=3x+2$ is continuous at $x=3$.

$f(x)=\dfrac{1}{x-2}$ is not defined when $x-2=0$, so the point of discontinuity is $x=2$.

(a) $f(1.99)=4(1.99)+7=14.96$ and $f(1.999)=4(1.999)+7=14.996$. As $x$ approaches $2$ from the left the values approach $15$, so

$\lim_{x\to2^-}f(x)=15.$

(b) $f(2.1)=4(2.1)+7=15.4$ and $f(2.01)=4(2.01)+7=15.04$. As $x$ approaches $2$ from the right the values approach $15$, so

$\lim_{x\to2^+}f(x)=15.$

(c) $f(2)=4(2)+7=15$. Since $\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=f(2)=15$, the function is continuous at $x=2$.

(d) Discontinuous, because continuity requires the functional value, the left-hand limit and the right-hand limit should be equal.

Here $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$, so the limit exists, but $f(a)$ is not defined. Since continuity needs $f(a)$ to exist and equal that limit, the function is discontinuous at $x=a$ (a removable discontinuity).

(a) A function $f(x)$ is continuous at $x=a$ if
$\lim_{x\to a^-} f(x)=\lim_{x\to a^+} f(x)=f(a).$
That is, the left-hand limit, right-hand limit and the value of the function at that point are all equal.

(b)
Left-hand limit at $x=3$: $\lim_{x\to 3^-}p(x)=\lim_{x\to 3}(2x+1)=2(3)+1=7$ Right-hand limit at $x=3$: $\lim_{x\to 3^+}p(x)=\lim_{x\to 3}(x+4)=3+4=7$ Since both limits are equal, $\lim_{x\to 3}p(x)=7$

(c)
Since $x=3$ lies in the interval $1\le x\le 3$, $p(3)=2(3)+1=7$

(d)
We have $\lim_{x\to 3}p(x)=7 \quad \text{and} \quad p(3)=7.$ So, $\lim_{x\to 3}p(x)=p(3).$ Therefore, the function $p(x)$ is continuous at $x=3$.