Given \(g(x)= \begin{cases}x^{2}-b^{2} & \text { if } x<4 \\ b x+20 & \text { if } x \geq 4\end{cases}\)
\(lim_{x\rightarrow 4^-}g(x)=lim_{x\rightarrow 4^-}x^2-b^2=(4)^2-b^2=16-b^2\)..(1)
\(lim_{x\rightarrow 4^+}g(x)=lim_{x\rightarrow 4^+}bx+20=b(4)+20=4b+20\)...(2)
Also, g(4) = bc + 20 = 4b + 20..(3)
Since g(x)is continuous on \((-\infty,\infty)\)
\(\Rightarrow lim_{x\rightarrow 4^-}g(x)=lim_{x\rightarrow 4^+}g(x)=g(4)\)
From (1), (2) and (3) we get
16-b² = 4b + 20
\(\Rightarrow b^2+4b+20-16=0\)
\(\Rightarrow b^2+4b+4=0\)
\(\Rightarrow (b+2)^2=0\)
\(\Rightarrow (b+2)=0\)
\(\Rightarrow b=-2\)

\(lim_{x\rightarrow\infty}{1^2+2^2+....+(3n)^2\over (1+2+...+5n)(2n+3)}\)

\([\because \sum n={n(n+1)\over 2} \ and \ \sum n^2 ={n(n+1)(2n+1)\over6}]\)

\(={1\over 5}.lim_{1\over n\rightarrow 0}{(3+{1\over n})(6+{1\over n})\over (5+{1\over n})(2+{3\over n})}\)

Hence proved.

Given \(f(x)={3+\sqrt{x}\over 9-x}\)
The curve does not exist atx = 9. Hence f(x) has a removable discontinuity atx = 9.
\(lim_{x\rightarrow 9}f(x)=lim_{x\rightarrow 9}{3-\sqrt{x}\over 9-x}\)

\(=lim_{x\rightarrow 9}{1\over (3-\sqrt{x})}={1\over 3+\sqrt{9}}={1\over 3+3}={1\over 6}\)
After removing the discontinuous point, the continuous functions g(x) can be written as
g(x) = {\(\begin{matrix} \frac { 3-\sqrt { x } }{ 9-x } & if\ x\ \neq 9 \\ \frac { 1 }{ 6 } & if\ x=9 \end{matrix}\)

Given \(f(x)={x^2-2x-8\over x+2}\)|
f(x) does not exist at x = - 2
\(\therefore\) It has a removable discontinuity at x = - 2
\(lim_{x\rightarrow -2}f(x)=lim_{x\rightarrow -2}{x^2-2x-8\over x+2}\)
\(lim_{x\rightarrow -2}f(x)=lim_{x\rightarrow -2}{(x-4)(x+2)\over (x+2)}=lim_{x\rightarrow-2}x-4=-2-4=-6\)
\(\therefore\) The continuous function g(x) can be written as
g(x) = {\(\begin{matrix} \frac { { x }^{ 2 }-2x-8 }{ x+2 } & if\ x\ \neq -2 \\ 6& if\quad x=-2 \end{matrix}\)

Given \(f(x)= \begin{cases}0 & \text { for } \quad x<0 \\ x & \text { for } \quad 0 \leq x<1 \\ -x^{2}+4 x-2 & \text { for } \quad 1 \leq x<3 \\ 4-x & \text { for } \quad x \geq 3\end{cases}\)
(i) At the point x = 0
\(lim_{x\rightarrow 0^-}f(x)=lim_{x\rightarrow 0^-}0=0\)
\(lim_{x\rightarrow 0^+}f(x)=x=0\)
and f(0) = x = 0
\(lim_{x\rightarrow 0^-}f(x)=lim_{x\rightarrow 0^+}f(x)=f(0)=0\)
\(\therefore f(x)\) is continuous at x = 0
(ii) At the point x = 1
\(lim_{x\rightarrow 1^-}f(x)=x=1\)
\(lim_{x\rightarrow 1^+}f(x)=-x^2+4x-2=-1+4-2=4-3=1\)
 f(1)=\(-x^2+4x-2=-1+4-2= 1\)
\(lim_{x\rightarrow 1^-}f(x)=lim_{x\rightarrow 1^+}f(x)=f(1)=1\)
\(\therefore f(x)\) is continuous at x = 1
(iii) At the point x = 3
\(lim_{x\rightarrow 3^-}f(x)=lim_{x\rightarrow 3^-}-x^2+4x-2=-(3)^2+4(3)-2\)
= -9 + 12 - 2 = 1
\(and \ lim_{x\rightarrow 3^+}f(x)=lim_{x\rightarrow 3^+}4-x=4-3=1\)
 f(3) = 4 - 3 = 1
\(\therefore lim_{x\rightarrow 3^-}f(x)=lim_{x\rightarrow 3^+}f(x)=f(3)=1\)
\(\therefore f(x)\) is continuous at x = 3

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.

\(limx_{x\rightarrow \infty}x[{3^{1\over x}+1-cos({1\over x}) -e^{1\over x}}]\)
Put \({1\over x}=y\)
\(\Rightarrow lim_{y\rightarrow 0}[3^y+1cos \ y-e^y]\) \(=limx_{y\rightarrow 0}{3^y-1+1+1-cos \ y-e^y\over y}\)  [Adding and subtracting 1 in the numerator]
\(=lim_{y\rightarrow 0}{3y-1\over y}-lim_{y\rightarrow0}({e^y-1\over y})+lim_{y\rightarrow0}{1-cos \ y \over y}\times {y \over y}\)
\(=log3-1+lim_{y\rightarrow 0}{2ysin^2{y\over 2}\over ({y^2\over 4}\times 4)}=log3-1+lim_{y\rightarrow 0}{y\over 2}.({sin{y\over 2}\over{y\over 2}})^2\)
= log 3-1 + 0 = log 3-1   \(\therefore lim_{x\rightarrow \infty}x[3^{1\over x}+1-cos({1\over x})-e^{1\over x}]\)
= log 3-1

\(lim_{x\rightarrow 0}{\sqrt{x^2+a^2}-a\over \sqrt{x^2+b^2}-b}\)\(=lim_{x\rightarrow 0}{({x^2+a^2})^{1\over2}-(a^2)^{1\over2}\over ({x^2+b^2})^{1\over2}-(b^2)^{1\over2}}\)\(=lim_{x\rightarrow 0}{({x^2+a^2})^{1\over2}-(a^2)^{1\over2}\over x^2}\times {x^2\over ({x^2+b^2})^{1\over2}-(b^2)^{1\over2}}\)
[Multiplying and dividing by x²]
\(=lim_{x\rightarrow 0}{({x^2+a^2})^{1\over2}-(a^2)^{1\over2}\over x^2+a^2-a^2}\times {x^2+b^2-b^2\over ({x^2+b^2})^{1\over2}-(b^2)^{1\over2}}\)
\(=[lim_{x^2+a^2\rightarrow a^2}{({x^2+a^2})^{1\over2}-(a^2)^{1\over2}\over x^2+a^2-a^2}]\times {1\over [lim_{x^2+b^2\rightarrow b^2} {({x^2+b^2})^{1\over2}-(b^2)^{1\over2}\over (x^2+b^2)-b^2}]}\)
\(={1\over2}(a^2)^{{1\over2}-1}\times {1\over {1\over2}(b^2)^{{1\over2}-1}}\)     \([\because lim_{x\rightarrow a}{x^n-a^n\over x-a}=n.a^{n-1}]\)
\(=(a^2)^{-{1\over2}}\times {1\over (b^2)^{1{1\over2}}}={\sqrt{b^2}\over \sqrt{a^2}}={b \over a}\)
\(\therefore lim_{x\rightarrow}{\sqrt{x^2+a^2}-a\over \sqrt{x^2+b^2}-b}={b\over a}\)

\({3^x-1\over \sqrt{1+x}-1}={(3^x-1)\over \sqrt{1+x}-1}{ \sqrt{1+x}+1\over \sqrt{1+x}+1}={(3^x-1)( \sqrt{1+x}+1)\over (1+x)-1}\)\(={(3^x-1)( \sqrt{1+x}+1)\over x}\)
Therefore \(lim_{x \rightarrow 0}{3^x-1\over \sqrt{1+x}-1}=lim_{x \rightarrow 0}{3^x-1\over x}.lim_{x\rightarrow0}\sqrt{1+x}+1\) = (log 3)(2) = 2 log 3 = log 9.