Calculus Solutions Ex#1.6

  Q.01 Find the discontinuities if any Solution: $\quad f(x)=\sin \left(x^{2}-2\right)$ This a composition of two functions. Let $$g(x)=x^{2}-2$$ and $$h(x)=\sin x$$ so we can write $$f(x)=h(g(x))$$ $g(x)$ is a polynomial which is continuous everywhere, and $h(x)=\sin x$ is continuous on $(-\infty,+\infty)$. Therefore, their composition is continuous ie $f(x)$ is continuous everywhere. There is no … Read more

Calculus Solutions Ex#1.5

Q.05 Solution: Consider the function $$f(x)=\left\{\begin{array}{ll} 1, & x \neq 4 \\ -1 & x=4 \end{array} \text { and } g(x)= \begin{cases}4 x-10, & x \neq 4 \\ -6 & x=4\end{cases}\right.$$ For each part, is the given function continuous at $x=4$ ?(a) $f(x)$ $$\begin{aligned} &f(x)=\left\{\begin{array}{ll} 1, & x \neq 4 \\ -1 & x=4 \end{array} … Read more

Calculus Solutions Ex#1.3

Q.09 Solution: $$\lim_{x \rightarrow+\infty}\left(1+2 x-3 x^5\right)=\lim_{x \rightarrow+\infty}\left(-3 x^5\right)=-\infty$$ This is because the behavior of a polynomial matches the end behavior of its highest degree term. Q.10 Solution: $$\lim _{x \rightarrow+\infty}\left(2 x^3-100 x+5\right)=\lim_{x \rightarrow+\infty} 2 x^3=+\infty.$$ $ \quad \left(2 x^3\right.$ is the highest degree term). Q.11 Solution: $$\quad \lim_{x \rightarrow+\infty} \sqrt{x}=\sqrt{\lim_{x \rightarrow+\infty} x}=+\infty$$. Q.12 Solution: $$\lim … Read more

Calculus Solutions Ex#1.2

Q.01 Solution: Given that $$\lim_{x \rightarrow a} f(x)=2,$$ $$ \lim_{x \rightarrow a} g(x)=-4$$ $$\lim_{x \rightarrow a} h(x)=0.$$(a) $$\begin{aligned} \lim_{x \rightarrow a}[f(x)+2 g(x)] & =\lim_{x \rightarrow a} f(x)+2 \underset{x \rightarrow a}{\operatorname{lim}} g(x) \\ & =2+2(-4)=2-8=-6 \end{aligned}$$(b) $$\begin{aligned} \lim _{x \rightarrow a}[h(x)-3 g(x)+1] & =\lim_{x \rightarrow a} h(x)-3 \underset{x \rightarrow a}{\operatorname{lim}} g(x)+\underset{x \rightarrow a}{\operatorname{lim}}(1) \\ & =0-3(-4)+1=13 … Read more

Calculus Solution Exercise 1.1

Q.01 For the function g graphed in …. find Solution Solution: (a) $\operatorname{lim}_{x \rightarrow 0^{-}} g(x)=3$ This is because as $x$ approaches o from the left side,  $g(x)$ approaches 3. (b) $\quad \lim _{x \rightarrow 0^{+}} g(x)=3$ From the graph, we can see that as $x$ approaches o from the right side,  $g(x)$ approaches 3. … Read more