\(lim_{x\rightarrow 0}{tan \ x-sin x\over x^3}\)\(=lim_{x\rightarrow 0}{{sin x\over cosx}-sin x\over x^3}\)\(=lim_{x\rightarrow 0}{sin x({1\over cosx}-1)\over x^3}\)
\(=(lim_{x\rightarrow 0}{sin x\over x})(lim_{x\rightarrow 0{1-cos x\over cos x(x^2)}})=1\times lim_{x\rightarrow 0} ({2sin^2{x\over2}\over cosx.{x^2\over 4}\times 4})\)
\(=lim_{x\rightarrow 0}{2\over 4cos x}\times lim_{{x\over 2}\rightarrow0}({sin^2{x\over 2}\over {x^4\over 4}})={1\over 2(1)}\times 1={1\over2}\)
\(\therefore lim_{x\rightarrow 0}{tan \ x-sin x\over x^3}={1\over2}\)

\(lim_{x\rightarrow 0}{\sqrt{2}-\sqrt{1+cos x}\over sin^2x}\)\(=lim_{x\rightarrow 0}{\sqrt{2}-\sqrt{2cos^2{x\over2}}\over sin^2x}\)\(=lim_{x\rightarrow 0}{\sqrt{2}-\sqrt{2}cos{x\over2}\over sin^2x}\) \([\because 1+cos \ x=2cos^2{x\over 2}]\)
\(=lim_{x\rightarrow 0}{\sqrt{2}(1-cos{x\over2})\over sin^2x}\)\(=\sqrt{2}lim_{x\rightarrow 0}{{2}sin^2({x\over4})\over sin^2x}\)
\(=2\sqrt{2}lim_{x\rightarrow 0}{{}sin^2({x\over4})\over ({x^2\over 16})}\times {{x^2\over 16}\over {sin^2x\over x^2}\times x^2}\)
\(=2\sqrt{2}.[lim_{{x\over 4}\rightarrow0}{sin({x\over4})\over ({x \over 4})}]^2\).\(({x^2\over 16\times x^2})\times {1\over(lim_{x\rightarrow 0}{sin x\over x})^2}\)
\(=2\sqrt{2}\times 1\times{1\over16}\times{1\over1}={2\sqrt{2}\over 16}={\sqrt{2}\over 8}\)

\(lim_{x\rightarrow 0}{sin \alpha x\over sin \beta x}\)\(=lim_{x\rightarrow 0}{sin \alpha x\over \alpha x }\times {\alpha x \over {sin \beta x\over \beta x}\times \beta x }=(lim_{\alpha \rightarrow0}{sin \alpha x\over \alpha x})\)\(\times (lim_{x\rightarrow0}{\alpha x \over \beta x})\times{1\over (lim_{\beta \times 0}{sin \beta x\over \beta x})}\)
\(=1\times {\alpha \over \beta }\times 1={\alpha \over \beta }\)  \([\because lim_{\theta \rightarrow 0}{sin \theta \over \theta}=1]\)

Given \(lim_{x\rightarrow3}[2f(x)-g(x)]=4\)
and f(3) = 5
\(\Rightarrow lim_{x\rightarrow3}2.f(x)-lim_{x\rightarrow3} g(x)=4\)
\(\Rightarrow 2.f(3)-g(3)=4\) [\(\therefore\) f(x) and g(x) are continuous function]
\(\Rightarrow 2(5)-g(3)=4\)
\(\Rightarrow 10-g(3)=4\)
\(\Rightarrow 10-4=g(3)\)
\(\Rightarrow g(3)=6\)

Given f(x) = {\(\begin{matrix} x+2, & if\quad x\ge 2 \\ { x }^{ 2 }, & if\quad x<2 \end{matrix}\)
\(lim_{x \rightarrow 2^-}f(x)=lim_{x\rightarrow2^-}x^2=2^2=4\)
\(lim_{x \rightarrow 2^+}f(x)=lim_{x\rightarrow2^+}x+2=2+2=4\)
Also f(2) =  x+2 = 2+2 = 4
\(\therefore lim_{x\rightarrow 2^-}f(x)=lim_{x\rightarrow 2^+}f(x)=f(2)=4\)
\(\therefore\) f(x)is continuous in R.

Given f(x) = {\(\begin{matrix} 4x+5, & if\quad x\le 3 \\ 4x-5, & if\quad x>3 \end{matrix}\)
\(lim_{x\rightarrow 3^-}f(x)=lim_{x\rightarrow 3^-}(4x+5)=4(3)+5=17\)
\(lim_{x\rightarrow 3^+}f(x)=lim_{x\rightarrow 3^+}(4x-5)=12-5=7\)
Since the \(lim_{x\rightarrow 3^-}f(x)\neq lim_{x\rightarrow 3^+}f(x)\)
f(x) is not continuous at x = 3.

The function \(y={1\over x}\) is continuous at all points of R except at x = 0 where it is undefined.
The function g(x) = sin \({1\over x}\) is continuous at all points except x = 0, where \(lim_{x \rightarrow 0}g(x)\) does not exist. So, g is continuous on the intervals \((-\alpha ,0)\) and \((0,\alpha )\)

\(lim_{x\rightarrow {\pi\over 2}}(1+sin x)^{2cosec \ x}\)\(=lim_{x\rightarrow {0}}(1+{1\over cosec \ x})^{2cosec \ x}\)
Put y = cosec x
when x\(\rightarrow\) 0, y\(\rightarrow\) cosec 0  [when x\(\rightarrow\) 0, cosec 0 \(\Rightarrow\) cosec x\(\rightarrow \infty\) ]
when x\(\rightarrow\)0, y \(\rightarrow \infty\)
\(=[lim_{y\rightarrow \infty}(1+{1\over y})^y]^2=e^2\) \([\because lim_{y\rightarrow \infty}(1+{1\over y})^y=e]\)

\(lim_{x \rightarrow \infty}\{ x[log(x+a)-log(x)]\}\)\(lim_{x\rightarrow \infty}x.log({x+a\over a})=lim_{x\rightarrow }xlog(1+{a\over x})\)
\(=lim_{x\rightarrow \infty }{log(1+{a\over x})\times a\over {1\over x}\times a}\)
Put \({1\over x}\times y\)
\(a.lim_{y\rightarrow0}{log(1+y)\over y}=a(1)=a\) \([\because lim_{x\rightarrow 0}log{(1+x)\over x}=1]\)

\(lim_{x\rightarrow 0}{2^x-3^x\over x}=lim_{x\rightarrow0}{2^x-1-3^x+1\over x}\) [Adding and subtracting 1 in the numerator]
\(lim_{x\rightarrow 0}{2^x-1\over x}-{(3^x+1)\over x}=lim_{x\rightarrow0}{2^x-1\over x}-lim_{x\rightarrow0}({3^x-1\over x})\)
\(=log2-log3\) \([\because lim_{x\rightarrow 0}{a^x-1\over x}=log \ a]\)
\(=log({2\over3})\)
\(\therefore lim_{x\rightarrow 0}{2^x-3^x\over x}=log({2\over3})\)