Exercise 5.1 Solutions

$$\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x – 3) = 5(0) – 3 = -3$$ $$f(0) = 5(0) – 3 = -3$$ $$\therefore \lim_{x \to 0} f(x) = f(0)$$

$$\lim_{x \to -3} f(x) = \lim_{x \to -3} (5x – 3) = 5(-3) – 3 = -18$$ $$f(-3) = 5(-3) – 3 = -18$$ $$\therefore \lim_{x \to -3} f(x) = f(-3)$$

$$\lim_{x \to 5} f(x) = \lim_{x \to 5} (5x – 3) = 5(5) – 3 = 22$$ $$f(5) = 5(5) – 3 = 22$$ $$\therefore \lim_{x \to 5} f(x) = f(5)$$

$$\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 – 1) = 2(3)^2 – 1 = 17$$ $$f(3) = 2(3)^2 – 1 = 17$$

(a) $f(x) = x – 5$
It is a polynomial function. It is defined and continuous for all real numbers.

(b) $f(x) = \frac{1}{x-5}$
The function is undefined at $x=5$. Hence, it is continuous for all $x \in \mathbb{R} – \{5\}$.

(c) $f(x) = \frac{x^2 – 25}{x+5}$
Defined for $x \neq -5$.
For $x \neq -5, f(x) = \frac{(x-5)(x+5)}{x+5} = x – 5$, which is continuous.
So, continuous at all points except $x = -5$.

(d) $f(x) = |x-5|$
Modulus functions are continuous everywhere. Thus, continuous for all $x \in \mathbb{R}$.

$$\lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n$$ $$f(n) = n^n$$ $$\text{Since } \lim_{x \to n} f(x) = f(n), \text{ the function is continuous.}$$

$$\text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1} x = 1$$ $$\text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1} 5 = 5$$ $$\text{LHL} \neq \text{RHL}$$

$$\text{LHL} = \lim_{x \to 2^-} (2x + 3) = 2(2) + 3 = 7$$ $$\text{RHL} = \lim_{x \to 2^+} (2x – 3) = 2(2) – 3 = 1$$ $$\text{Since LHL} \neq \text{RHL, f is discontinuous at } x = 2$$

$$\text{LHL} = \lim_{x \to -3^-} (|x|+3) = |-3|+3 = 6$$ $$\text{RHL} = \lim_{x \to -3^+} (-2x) = -2(-3) = 6$$ $$f(-3) = |-3|+3 = 6$$ $$\text{Continuous at } x = -3$$

$$\text{LHL} = \lim_{x \to 3^-} (-2x) = -2(3) = -6$$ $$\text{RHL} = \lim_{x \to 3^+} (6x+2) = 6(3)+2 = 20$$ $$\text{LHL} \neq \text{RHL}$$

$$\text{LHL} = \lim_{x \to 0^-} \frac{-x}{x} = -1$$ $$\text{RHL} = \lim_{x \to 0^+} \frac{x}{x} = 1$$ $$\text{LHL} \neq \text{RHL}$$

$$\text{LHL} = \lim_{x \to 0^-} \frac{x}{-x} = -1$$ $$\text{RHL} = \lim_{x \to 0^+} (-1) = -1$$ $$f(0) = -1$$

$$\text{LHL} = \lim_{x \to 1^-} (x^2+1) = 1^2+1 = 2$$ $$\text{RHL} = \lim_{x \to 1^+} (x+1) = 1+1 = 2$$ $$f(1) = 1+1 = 2$$

$$\text{LHL} = \lim_{x \to 2^-} (x^3 – 2) = 2^3 – 2 = 6$$ $$\text{RHL} = \lim_{x \to 2^+} (x + 1) = 2 + 1 = 3$$ $$\text{LHL} \neq \text{RHL}$$

$$\text{LHL} = \lim_{x \to 1^-} (x^2 – 1) = 1^2 – 1 = 0$$ $$\text{RHL} = \lim_{x \to 1^+} x = 1$$ $$\text{LHL} \neq \text{RHL}$$

$$\text{LHL} = \lim_{x \to 1^-} (x+5) = 1+5 = 6$$ $$\text{RHL} = \lim_{x \to 1^+} (x-5) = 1-5 = -4$$ $$\text{LHL} \neq \text{RHL}$$

Q14: Check points $x=1, 3$.
At $x=1$: LHL=3, RHL=4 $\implies$ Discontinuous.
At $x=3$: LHL=4, RHL=5 $\implies$ Discontinuous.

Q15: Check points $x=0, 1$.
At $x=0$: LHL=0, RHL=0, f(0)=0 $\implies$ Continuous.
At $x=1$: LHL=0, RHL=2 $\implies$ Discontinuous.

Q16: Check points $x=-1, 1$.
At $x=-1$: LHL=-2, RHL=-2, f(-1)=-2 $\implies$ Continuous.
At $x=1$: LHL=2, RHL=2, f(1)=2 $\implies$ Continuous.

$$\text{LHL} = \lim_{x \to 3^-} (ax+1) = 3a+1$$ $$\text{RHL} = \lim_{x \to 3^+} (bx+3) = 3b+3$$ $$\text{Since continuous: } 3a+1 = 3b+3$$ $$3a – 3b = 2 \implies a – b = \frac{2}{3}$$

$$\text{LHL} = \lim_{x \to 0^-} \lambda(x^2 – 2x) = \lambda(0) = 0$$ $$\text{RHL} = \lim_{x \to 0^+} (4x+1) = 1$$ $$\text{LHL} \neq \text{RHL for any value of } \lambda.$$

$$\text{LHL} = \lim_{x \to c^-} (x – [x]) = c – (c-1) = 1$$ $$\text{RHL} = \lim_{x \to c^+} (x – [x]) = c – c = 0$$ $$\text{LHL} \neq \text{RHL}$$

Q20: $f(x) = x^2 – \sin x + 5$
Sum of continuous functions ($x^2$, $\sin x$, constant) is continuous. $\therefore$ Continuous at $x=\pi$.

Q21: $\sin x + \cos x$, $\sin x – \cos x$, $\sin x \cdot \cos x$
Since $\sin x$ and $\cos x$ are continuous everywhere, their sum, difference, and product are also continuous everywhere.

Q23: Find discontinuity for $f(x) = \begin{cases} \frac{\sin x}{x}, & x < 0 \\ x+1, & x \ge 0 \end{cases}$
At $x=0$: LHL $= \lim_{x \to 0^-} \frac{\sin x}{x} = 1$. RHL $= 0+1=1$. f(0)=1.
Continuous everywhere.

Q24: $f(x) = \begin{cases} x^2 \sin(1/x), & x \neq 0 \\ 0, & x=0 \end{cases}$
Limit as $x \to 0$ is $0 \times (\text{value between -1 and 1}) = 0$. Matches $f(0)$. Continuous.

Q25: $f(x) = \begin{cases} \sin x – \cos x, & x \neq 0 \\ -1, & x=0 \end{cases}$
Limit as $x \to 0$ is $\sin 0 – \cos 0 = -1$. Matches $f(0)$. Continuous.

Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}, h \to 0$. $$\lim_{h \to 0} \frac{k \cos(\frac{\pi}{2}+h)}{\pi – 2(\frac{\pi}{2}+h)} = \lim_{h \to 0} \frac{k(-\sin h)}{-2h}$$ $$= \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} = \frac{k}{2}(1) = \frac{k}{2}$$ $$\text{For continuity: } \frac{k}{2} = 3 \implies k = 6$$

Q27: At $x=2$: $k(2)^2 = 3 \implies 4k=3 \implies k = \frac{3}{4}$.

Q28: At $x=\pi$: $k\pi + 1 = \cos \pi = -1 \implies k\pi = -2 \implies k = \frac{-2}{\pi}$.

Q29: At $x=5$: $k(5) + 1 = 3(5) – 5 \implies 5k + 1 = 10 \implies 5k = 9 \implies k = \frac{9}{5}$.

$$2a + b = 5 \quad \dots(1)$$

$$10a + b = 21 \quad \dots(2)$$

Subtract (1) from (2): $8a = 16 \implies a = 2$ Substitute $a=2$ in (1): $2(2) + b = 5 \implies 4 + b = 5 \implies b = 1$

Q31: $f(x) = \cos(x^2)$. Composite of $g(x)=\cos x$ and $h(x)=x^2$. Both continuous $\implies$ Composition is continuous.

Q32: $f(x) = |\cos x|$. Composite of modulus and cosine. Continuous.

Q33: $f(x) = \sin|x|$. Composite of sine and modulus. Continuous.

Q34: $f(x) = |x| – |x+1|$. Difference of two continuous modulus functions. Hence, continuous for all real numbers. No points of discontinuity.