From the graph of \(f(x)={1\over (x-2)^3},\)

clearly,\({1\over (x-2)^3}\rightarrow - \infty \ as \ x\rightarrow 2^- \)  and \({1\over (x-2)^3}\rightarrow \infty \ as \ x\rightarrow 2^+\)
Hence the limit does not exist

One can tabulate values of x near 0 (from either side) and conclude \(f(x)={1\over x^2+x^3}\) grows without bound and hence \(f(x)\rightarrow \infty as \ x\rightarrow 0.\)
To calculate this limit without making a table, we first divide the numerator and denominator by x². This division can be done, since in the calculation of the limit x ≠ 0 and hence x² ≠ 0. We can have
\(lim_{x\rightarrow 0}{1\over x^2+x^3}=lim_{x\rightarrow 0}{{1\over x^2}\over{x^2+x^3\over x^2}}={lim_{x\rightarrow}({1\over x^2})\over lim_{x\rightarrow0}(1+x)}\)
Now \({1\over 8}\rightarrow \infty as \ x\rightarrow 0\) and \(lim_{x\rightarrow0}(1+x)=1\) .
Thus the numerator grows without bound while the denominator approaches 1, implying that \({1\over (x^2+x^3)}\) does tend to infinity.

\(={-1\over 2(2)}=-{1\over4}\)

\(lim_{x\rightarrow3^-}f(x)=9a-4b+1\)
\(lim_{x\rightarrow3^+}f(x)=3a+b.\) Now the existence of limit forces us to have
\(lim_{x\rightarrow3^-}f(x)=lim_{x\rightarrow3^+}f(x)\).
\(\Rightarrow \) 9a - 4b + 1 = 3a + b
\(\Rightarrow \)6a - 5b + 1 = 0.

We can’t apply the quotient theorem immediately. Use the algebra technique of rationalising the numerator
\({\sqrt{t^2+9}-3\over t^2}={({\sqrt{t^2+9}-3})({\sqrt{t^2+9}-3})\over t^2({\sqrt{t^2+9}-3})}={t^2+9-9\over t^2({\sqrt{t^2+9}-3})}\)
\(lim_{t\rightarrow0}{\sqrt{t^2+9}-3\over t^2}=lim_{t\rightarrow0}{t^2\over t^2\sqrt{t^2+9+3}}=lim_{t\rightarrow o}{1\over\sqrt{t^2+9+3} }\)\(={1\over \sqrt{9}+3}={1\over6}\).

\(lim_{x\rightarrow3}(x^2-6x+5)=3^2-6\times 3+5=-4\)
\(lim_{x\rightarrow3}(x^3-8x+7)=3^2-8\times 3+7=10\neq0.\)
Therefore,\(lim_{x\rightarrow3}{(x^2-6x+5)\over x^3-8x+7}={lim_{x\rightarrow3}(x^2-6x+5)\over lim_{x\rightarrow3}(x^3-8x+7)}={-4\over 10}=-{2\over5}\) .

Given\(f(x)= \begin{cases}\frac{|x-1|}{x-1}, & \text { for } x \neq 1 \\ 0, & \text { for } x=1\end{cases}\)
\(f(1^-)=lim_{x\rightarrow 1^-}{|x-1|\over x-1}=lim_{x\rightarrow 1^-}{-(x-1)\over x-1}=-1\)
\(f(1^+)=lim_{x\rightarrow 1^+}{|x-1|\over x-1}=lim_{x\rightarrow 1^+}{x-1\over x-1}=1\)
Since f(1^-) \(\neq\) f (1⁺), the limit of f(x) does not exist.

f(3^-)=\(lim_{x\rightarrow{3^-}}{x^2-9\over x-3}\)

\(=lim_{x\rightarrow{3^-}}(x+3)=3+3=6\)
f(3⁺)=\(lim_{x\rightarrow{3^+}}{x^2-9\over x-3}\)

\(=lim_{x\rightarrow{3^+}}(x+3)=6\)

Given \(f(x)=tan \ x \ at \ x={\pi\over 2}.\)
lim f(x) = lim tan x = \(\infty\)
\( x\rightarrow{\pi^-\over 2}\)   \( x\rightarrow{\pi^-\over 2}\)
\(\therefore lim_{ x\rightarrow{\pi^-\over 2}} f(x)\rightarrow \infty \ as \ x\rightarrow{\pi^-\over 2}\)
lim f(x) = lim tan x
\( x\rightarrow{\pi^+\over 2}\)   \( x\rightarrow{\pi^+\over 2}\)
= \(-\infty\)     [\(\because \ tan \pi =tan (90+90)=-cot 90=-\infty\)]
\(\therefore lim_{x\rightarrow{\pi^+\over 2}} f(x)\rightarrow -\infty \ as \ x\rightarrow{\pi^+\over 2}\)

\(lim_{\sqrt x\rightarrow3}{x^2-81\over \sqrt x-3}\)=\(lim_{\sqrt x\rightarrow3}{(\sqrt x)^4-3^4\over \sqrt x-3}\) \([\because lim_{x\rightarrow a}{x^n-a^n\over x-a}=n.a^{n-1}]\)
= 4.3^4-1 = 4(3)³
= 4(27) = 108