\(lim_{x\rightarrow 1^-}f(x)=lim_{x\rightarrow 1^-}3x=3\)
\(lim_{x\rightarrow 1^+}f(x)=lim_{x\rightarrow 1^+}2x-1=2-1=1\)
\(lim_{x\rightarrow 1^-}f(x)\neq lim_{x\rightarrow 1^+}f(x)\)
and \(f(1)=2x-1=2(1)-1=1\)
\(\therefore f(x)\) is not continuous at x = 1

Given \(f(x)= \begin{cases}0, & \text { if } x<0 \\ x^{2}, & \text { if } 0 \leq x<2 \\ 4, & \text { if } x \geq 2\end{cases}\)
\(lim_{x\rightarrow 2^-}f(x)=lim_{x\rightarrow 2^-}x^2=4\)
\(lim_{x\rightarrow 2^+}f(x)=lim_{x\rightarrow 2^+}4=4\)
Also f(2) = 4
\(\therefore lim_{x\rightarrow 2^-}f(x)=lim_{x\rightarrow 2^+}f(x)=f(2)=4\)
\(\therefore f(x)\) is continuous on \((-\infty,\infty)\)

Let x denote the number of kilograms bought per day and C denote the cost. Then,
\(C(x)= \begin{cases}0.16 x, & \text { if } 0 \leq x<100 \\ 0.14 x, & \text { if } x \geq 100\end{cases}\)
The sketch of this function
It is discontinuous at x = 100 since \(lim_{x\rightarrow100^-}c(x)=16\) and \(lim_{x\rightarrow100^+}c(x)=14\)
Note that C(100) = 14. Thus,\(lim_{x\rightarrow100^-}c(x)=16\neq 14=lim_{x\rightarrow100^+}C(x)=C(100).\)
Note also that the function jumps from one finite value 14 to another finite value 16.

The function h(x) is defined at all points of the real line R \(=(-\infty,\infty);\) for any \(x_o \neq 0,\)
\(lim_{x\rightarrow x_o}h(x)=(lim_{x\rightarrow x_O}xsin{1\over x})\)
\(=x_o sin {1\over x_o}=h(x_o)\)
For x₀ = 0
\(h(x)=x.sin{1\over x}\)
\(-x\le xsin{1\over x}\le x\)
\(g(x)=-x,f(x)=xsin{1\over x},h(x)=x\)
\(lim_{x\rightarrow 0}g(x)=0,lim_{x\rightarrow 0}h(x)=0\)
and have\(lim_{x\rightarrow 0}x \ sin {1\over x}=0.\)
By Sandwich theorem
\(lim_{x\rightarrow 0}(x \ sin {1\over x})=0=h(0).\)
Therefore h(x) is continuous in the entire real line.

\(lim_{\alpha\rightarrow 0}{sin (\alpha)^n\over( sin \alpha )^m}\)
Case (i): If m=n,
\(lim_{\alpha\rightarrow 0}{sin (\alpha)^n\over( sin \alpha )^m}\)=\(lim_{\alpha \rightarrow 0}{sin(\alpha^m)\over \alpha^m}\times{\alpha^m\over (sin \alpha )^m}\)
\(=(lim_{a^m\rightarrow 0}{sin(\alpha^m)\over (\alpha)^m})\times{1\over lim_{\alpha \rightarrow0({sin \alpha \over \alpha})^m}}=1\times {1\over 1^m}=1[\because lim_{\theta \rightarrow 0}{sin \theta\over \theta}=1]\)
\(\therefore lim_{\alpha\rightarrow 0}{sin (\alpha)^n\over( sin \alpha )^m}=1\)
Case (ii): If m>n,
\(lim_{\alpha\rightarrow 0}{sin (\alpha)^n\over( sin \alpha )^m}=lim_{\alpha \rightarrow 0}{sin(\alpha^n)\over \alpha^n}\times {\alpha^n\over {(sin \alpha)^m.\alpha^m\over \alpha^m}}\)\(=({lim_{\alpha^n\rightarrow 0}{sin(\alpha^n)\over \alpha^n}}).{\alpha^n\over \alpha^n}.{1\over (lim_{\alpha\rightarrow0}{sin \alpha \over \alpha})^m}\)
\(=1\times \alpha^{n-m}\times 1=\alpha^{n-m}=0since \ m>n\)
Case (ii): If m \(lim_{\alpha\rightarrow 0}{sin (\alpha)^n\over( sin \alpha )^m}=lim_{\alpha \rightarrow 0}{sin(\alpha^n)\over \alpha^n}\times {\alpha^n\over {(sin \alpha)^m\times\alpha^m\over \alpha^m}}\)
\(=({lim_{\alpha^n\rightarrow 0}{sin(\alpha^n)\over \alpha^n}})\times{\alpha^n\over \alpha^n}\times{1\over lim_{\alpha\rightarrow0}({sin \alpha \over \alpha})^m}\)\(=1\times \alpha^{n-m}\times1={1\over \alpha^{m-n}}\)
\(Since m

\(\frac{\sin (x-\lfloor x\rfloor)}{x-\lfloor x\rfloor}= \begin{cases}\frac{\sin (x-(-1))}{x-(-1)} & \text { if }-1<x<0 \\ \frac{\sin (x-0)}{x-0} & \text { if } 0<x<1\end{cases} \)
\(f(x)= \begin{cases}\frac{\sin (x+1)}{(x+1)} & \text { if }-1<x<0 \\ \frac{\sin x}{x} & \text { if } 0<x<1\end{cases}\)
\(lim_{x \rightarrow 0^-}f(x)={sin1\over1}=sin 1\)
\(lim_{x \rightarrow 0^+}f(x)=1\).
Hence the limit does not exist.

\(lim_{t\rightarrow \infty}r(t)=lim_{t\rightarrow \infty}-\sqrt{32\over k}{1-e^{2t\sqrt{32k}}\over1+e^{-2r\sqrt{32k}}}\)
\(=-\sqrt{32\over k}lim_{t\rightarrow \infty}{1-e^{2t\sqrt{32k}}\over1+e^{-2r\sqrt{32k}}}\)
\(=-\sqrt{32\over k}{(1-0)\over (1+0)}=-\sqrt{32\over k}ft/sec.\)

\({1\over x}-1\le \left\lfloor {1\over x} \right\rfloor \le{1\over x}+1\)
\({2\over x}-1\le \left\lfloor {2\over x} \right\rfloor \le{2\over x}+1\)
\({15\over x}-1\le \left\lfloor {15\over x} \right\rfloor \le{15\over x}+1\)
Summing, we get,
\({120\over x}-15 \le \left\lfloor {1\over x} \right\rfloor+\left\lfloor {2\over x} \right\rfloor +....+\left\lfloor {15\over x}\right\rfloor { 120\over x} +15\)
\({120}-15 \le x[ \left\lfloor {1\over x} \right\rfloor+\left\lfloor {2\over x} \right\rfloor +....+\left\lfloor {15\over x}\right\rfloor]\le { 120} +15x\)
\(lim_{x\rightarrow0^+}({120}-15 x)\le lim_{x\rightarrow0^+}x[ \left\lfloor {1\over x} \right\rfloor+\left\lfloor {2\over x} \right\rfloor +....+\left\lfloor {15\over x}\right\rfloor] \le lim_{x\rightarrow0^+}\le { (120} +15x)\)
\(120\le lim_{x\rightarrow0^+}x[ \left\lfloor {1\over x} \right\rfloor+\left\lfloor {2\over x} \right\rfloor +....+\left\lfloor {15\over x}\right\rfloor]\le120\)
\( lim_{x\rightarrow0^+}x[ \left\lfloor {1\over x} \right\rfloor+\left\lfloor {2\over x} \right\rfloor +....+\left\lfloor {15\over x}\right\rfloor]=120\).

LHS=\(lim_{x\rightarrow\infty} {1\over 1.2}+{1\over 2.3}+{1\over 3.4}+...+{1\over n(n+1)}=1\)
\(=lim_{n\rightarrow \infty}{6\over n(n+1)(2n+1)+3n(n+1)}=lim_{n\rightarrow \infty}{6\over n(n+1)[2n+1+3]}\)
\(=lim_{n\rightarrow \infty}{6\over n(n+1)[2n+4]}==lim_{n\rightarrow \infty}{3\over n(n+1)(n+2)}\)
\(=lim_{{{1\over n }\rightarrow 0}}{3\over (1+{1\over n})(1+{2\over n})}\)\(={3\over (1+0)(1+0)}\times{1\over n^3}=1\)

Consider \(lim_{n\rightarrow\infty}{1+2+3+...+n\over 3n^2+7n+2}\)
LHS \(=lim_{n\rightarrow \infty}{n(n+1)\over 2[3n^2+7n+2]}\)  \([\because \sum n={n(n+1)\over 2}]\)

\(={1\over2}lim_{{1\over n}\rightarrow 0}{(1+{1\over n})\over(3+{7\over n}+{2\over n^2})}\)
\(={1\over2}({1\over3})={1\over 6}=RHS\)
Hence proved.