Given \(f(x)={x^2-4\over (x^2+4x+4)(x+3)}at \ x=-2\)

\(=lim_{x\rightarrow -2^-}{x-2\over (x+2)(x+3)}\)
\(={Negative \over 0(Negative)}=\infty\)
\(\therefore f(-2^-)\rightarrow \infty as \ x\rightarrow 2^-\)
\(f(-2)^+=lim_{x\rightarrow -2^+}{(x+2)(x-2)\over (x+2)(x+3)}\)
\(=lim_{x\rightarrow -2^+}{x-2\over (x+2)(x+3)}\)
\(={Negative \over 0(Positive)}\rightarrow-\infty\)
\(\therefore f(-2^+)\rightarrow -\infty as \ x\rightarrow 2^+\).

\(lim_{x\rightarrow{1}}sin \pi x\)
At x = 1, the curve meets the x-axis.
\(\therefore lim_{x\rightarrow{1}}sin \pi x=0\)

\(|x|= \begin{cases}-x & \text { if } x<0 \\ 0 & \text { if } x=0 \\ x & \text { if } x>0\end{cases}\)
If x > 0,then |x| =  x, which tends to 0 as
\(x \rightarrow 0\) from the right of 0. That is, \(\lim _{ x\rightarrow0^+}{|x| } =0\)
If x < 0, then |x| = - x which again tends to 0 as x\(\rightarrow\)0. from the left of 0. That is, \(\lim _{ x\rightarrow0^-}{|x| } =0\).
Thus, \(\lim _{ x\rightarrow0^-}{|x| } =0=\lim _{ x\rightarrow0^+}{|x| }.\)
Hence \(\lim _{ x\rightarrow0}{|x| } =0\).

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

Given f(0)=2
\(\lim _{ x\rightarrow 0 }{ f\left( x \right) } =\lim _{ x\rightarrow 0 }{ \frac { \sin { 2x } }{ \sin { 3x } } } =\lim _{ x\rightarrow 0 }{ \left( \frac { \sin { 2x } }{ 2x } \right) } \left( \frac { 3x }{ \sin { 3x } } \right) \left( \frac { 2 }{ 3 } \right) \)
\(=\left( \lim _{ 2x\rightarrow 0 }{ \frac { \sin { 2x } }{ 2x } } \right) \left( \lim _{ 3x\rightarrow 0 }{ \frac { 1 }{ \frac { \sin { 3x } }{ 3x } } } \right) \times \frac { 2 }{ 3 } =1\times 1\times \frac { 2 }{ 3 } =\frac { 2 }{ 3 } \)
\(\therefore \lim _{ x\rightarrow 0 }{ f\left( x \right) } \neq f\left( 0 \right) \)

\(Put\quad x-\pi =\theta \Rightarrow x\quad =\pi +\theta \)
\(Also\quad \theta =x-\pi \rightarrow 0\quad as\quad x\rightarrow \pi \)
\(\lim _{ x\rightarrow \pi }{ \frac { \sin { x } }{ x-\pi } } =\quad \lim _{ \theta \rightarrow 0 }{ \frac { \sin { (\pi +\theta ) } }{ \theta } } = \lim _{ \theta \rightarrow 0 }{ \frac { -\sin { \theta } }{ \theta } } \)
\(=(-1).\lim _{ \theta \rightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } =(-1)(1)=-1\)

\(\lim _{ x\rightarrow 1 }{ \frac { (2x-3)\sqrt { x } -1 }{ { 2x }^{ 2 }+x-3 } } =\lim _{ x\rightarrow 1 }{ \frac { (2x-3)(\sqrt { x } -1)(\sqrt { x } +1) }{ (x-1){ (2x }+3)(\sqrt { x } +1) } } =\lim _{ x\rightarrow 1 }{ \frac { (2x-3)(x-1) }{ (x-1){ (2x }+3)(\sqrt { x } +1) } } \)
\(=\lim _{ x\rightarrow 1 }{ \frac { 2x-3 }{ { (2x }+3)(\sqrt { x } +1) } } =\frac { -1 }{ (2+3)(1+1) } \)
\(=\frac { -1 }{ 5(2) } =\frac { -1 }{ 10 } \)

\(=\frac { { 2 }^{ 2 }+2(2)+4 }{ 2+2 } =\frac { 12 }{ 4 } =3\)

\(lim_{x\rightarrow\infty}{x^4-5x\over x^2-3x+1 }\)\(=lim_{x\rightarrow\infty}{x^4(1-{5\over x^3})\over x^2(1-{3\over x}+{1\over x^2})}\)\(=lim_{x\rightarrow\infty}{x^2(1-{5\over x^3})\over(1-{3\over x}+{1\over x^2})}\)\([\therefore {1\over x}\rightarrow 0 \ as \ x \rightarrow \infty]\)
\(=\infty\)
\(lim_{x\rightarrow\infty}{x^4-5x\over x^2-3x+1 }=\infty\)