Differentiation (Part-38) | S N De

1–3. Piecewise Functions and One-Sided Limits

$1.$ The function $f(x)$ is defined as follows: $$f(x) = \begin{cases} 2-3x, & \text{when } x < 0 \\ 3x-2, & \text{when } x \ge 0 \end{cases}$$ Evaluate $\lim_{x \to 0+}f(x)$ and $\lim_{x \to 0-}f(x)$ and state whether $\lim_{x \to 0}f(x)$ exists or not.

Solution

Right Hand Limit (R.H.L.):

$$\lim_{x \to 0+}f(x) = \lim_{x \to 0+}(3x-2) \quad [\text{since } x \to 0^+ \implies x > 0]$$ $$= 3(0) - 2 = -2$$

Left Hand Limit (L.H.L.):

$$\lim_{x \to 0-}f(x) = \lim_{x \to 0-}(2-3x) \quad [\text{since } x \to 0^- \implies x < 0]$$ $$= 2 - 3(0) = 2$$

Since $\lim_{x \to 0+}f(x) \neq \lim_{x \to 0-}f(x)$,

Conclusion: $\lim_{x \to 0}f(x)$ does not exist

$2.$ If $\phi(x)$ is defined as: $$\phi(x) = \begin{cases} e^{2x}+1, & \text{when } x \ge 0 \\ 1+\cos 2x, & \text{when } x < 0 \end{cases}$$ Show that $\lim_{x \to 0}\phi(x) = 2$.

Solution

Right Hand Limit (R.H.L.):

$$\lim_{x \to 0+}\phi(x) = \lim_{x \to 0+}[e^{2x}+1] = e^{2(0)} + 1 = 1 + 1 = 2$$

Left Hand Limit (L.H.L.):

$$\lim_{x \to 0-}\phi(x) = \lim_{x \to 0-}[1+\cos 2x] = 1 + \cos(2 \cdot 0) = 1 + 1 = 2$$

Since $\lim_{x \to 0+}\phi(x) = \lim_{x \to 0-}\phi(x) = 2$,

$$\therefore \lim_{x \to 0}\phi(x) = 2 \quad \text{(Showed)}$$

$3.$ If $f(x) = [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$, show that $\lim_{x \to 2} f(x)$ does not exist.

Solution

By definition of the greatest integer function:

$$[x] = \begin{cases} 1, & \text{when } 1 \le x < 2 \\ 2, & \text{when } 2 \le x < 3 \end{cases}$$

Evaluating one-sided limits at $x = 2$:

$$\lim_{x \to 2+}f(x) = \lim_{x \to 2+}[x] = 2$$ $$\lim_{x \to 2-}f(x) = \lim_{x \to 2-}[x] = 1$$

Since $\lim_{x \to 2+}f(x) \neq \lim_{x \to 2-}f(x)$,

Conclusion: $\lim_{x \to 2}f(x)$ does not exist

4–5. Evaluation of Limits by Factorization

$4.$ Evaluate: $$\lim_{x \to 1}\frac{x^2-3x+2}{x^3-4x+3}$$

Solution

First, we factorize the numerator and denominator:

Numerator Factorization:

$$x^2-3x+2 = x^2-2x-x+2 = x(x-2)-1(x-2) = (x-2)(x-1) \quad \dots (1)$$

Denominator Factorization:

$$x^3-4x+3 = x^3-x^2+x^2-x-3x+3 = x^2(x-1)+x(x-1)-3(x-1) = (x-1)(x^2+x-3) \quad \dots (2)$$

Now, evaluating the limit:

$$\lim_{x \to 1}\frac{x^2-3x+2}{x^3-4x+3} = \lim_{x \to 1}\frac{(x-2)(x-1)}{(x-1)(x^2+x-3)}$$ $$= \lim_{x \to 1}\frac{x-2}{x^2+x-3} = \frac{1-2}{1^2+1-3} = \frac{-1}{-1} = 1 \quad \text{(Ans.)}$$

$5.$ Evaluate: $$\lim_{x \to -2}\frac{x^3-7x-6}{x^4+5x-6}$$

Solution

Factorizing the numerator and denominator expressions:

Numerator:

$$x^3-7x-6 = x^3+1-7x-7 = (x+1)(x^2-x+1)-7(x+1)$$ $$= (x+1)(x^2-x-6) = (x+1)(x-3)(x+2) \quad \dots (1)$$

Denominator:

$$x^4+5x-6 = x^4-1+5x-5 = (x^2+1)(x^2-1)+5(x-1)$$ $$= (x-1)[(x^2+1)(x+1)+5] = (x-1)(x^3+x^2+x+6)$$

Factorizing $(x^3+x^2+x+6)$:

$$x^3+2x^2-x^2-2x+3x+6 = x^2(x+2)-x(x+2)+3(x+2) = (x+2)(x^2-x+3)$$ $$\therefore x^4+5x-6 = (x-1)(x+2)(x^2-x+3) \quad \dots (2)$$

Evaluating the limit using $(1)$ and $(2)$:

$$\lim_{x \to -2}\frac{x^3-7x-6}{x^4+5x-6} = \lim_{x \to -2}\frac{(x+1)(x-3)(x+2)}{(x-1)(x+2)(x^2-x+3)}$$ $$= \lim_{x \to -2}\frac{(x+1)(x-3)}{(x-1)(x^2-x+3)}$$ $$= \frac{(-2+1)(-2-3)}{(-2-1)(4+2+3)} = \frac{(-1)(-5)}{(-3) \times 9} = -\frac{5}{27} \quad \text{(Ans.)}$$