The left-hand limit and right-hand limit does not coincide at x=x_o.

The limit of f(x) does not exist at x = x_o.

Given  f(x) = 2x² + 3x - 5
f(x) is an algebraic function.
Since the algebraic function is continous in R
f(x) is comtinous at all points in R.

By Sandwitch theorem \(lim_{x\rightarrow 0}x^2sin{1\over x}=0\) and f(0) = 0 by the definition of f(x). Hence it is continuous at x = 0. For other values it is clearly continuous and hence continuous in R.

\(lim_{x\rightarrow0}{\sqrt{1+x}-1\over x}\)
Multiplying and dividing by\((\sqrt{1+x}+1)\) we get,
\(lim_{x\rightarrow0}{\sqrt{1+x}-1\over x}\times {\sqrt{1+x}+1\over \sqrt{1+x}+1}= lim_{x\rightarrow0}{(1+x)-1\over x[\sqrt{1+x}+1]}\)

\(=lim_{x\rightarrow 0}{1\over \sqrt{1+x}+1}={1\over \sqrt{1}+1}={1\over1+1}={1\over2}\)

\(lim_{x\rightarrow1}{x^m-1\over x^n-1}\)
Multiplying and dividing by (x - 1) we get,
\(lim_{x\rightarrow1}{x^m-1^m\over x-1} \times { x-1\over x^n-1^n}=\)\((lim_{x\rightarrow1}{x^m-1^m\over x-1}) \times lim_{x\rightarrow1} {1\over ({x^n-1^n\over x-1})}\)
\(=m.(1)^{m-1}\times {1\over n(1)^{n-1}}={m \over n}\) \([\because lim_{x\rightarrow a}{x^n-a^n\over x-a}=n.a^{n-1}]\)

Here \(lim_{x\rightarrow1}(x-1)=0.\) In such cases, rationalise the numerator
\(lim_{x\rightarrow1}{\sqrt{x}-1\over x-1}=lim_{x\rightarrow1}{(\sqrt{x}-1)\over (\sqrt{x}-1)(\sqrt{x}+1)}=lim_{x\rightarrow{1}}{1\over\sqrt{x}+1} \)\(={{lim_{x\rightarrow1}}(1)\over lim_{x\rightarrow1}(\sqrt{x}+1)}={1\over2}\) .

\(lim_{x\rightarrow-2}(-{3\over 2}x)=-{3\over2}lim_{x\rightarrow-2}(x)=({-{3\over2}})(-2)=3.\)

Limit of f(x) as x approaches 2 is the nature of f(x) on both sides of 2.
It is independent of the nature of f(x) at x = 2.
Therefore we can not conclude anything about \(\lim _{x \rightarrow 2} \) from f(2) = 4.

\(\frac{1}{x-3}\) can be made arbitrarily large by choosing x suficiently close to 1 on the right side but not equal to 1.
\(\therefore \frac{1}{x-3}\) does not approach any value when x approaches 3 from the right.
\(\therefore \lim _{x \rightarrow 3^{+}} \frac{1}{x-3}=x\) and hence the limit does not exist.