Q.Prove that $f(x)=2x^2+3x-5$ is continuous at all points in $\mathbb{R}$.

$f(x)=1/x$ has domain $\mathbb{R}-\{0\}$ (undefined at $x=0$), and it is continuous at every point of this domain.

$f(x)=\dfrac{1}{x}$ is a rational function, defined for all $x\ne0$.

Rational functions are continuous at every point of their domain (quotient of continuous polynomials, valid wherever the denominator is nonzero).

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Cancel the common $(x-1)$ factor and take the limit; continuity forces $f(1)$ to equal that limit, $\dfrac23$.

Removable discontinuity and continuity at a point is a CBSE/NCERT Class 12 continuity topic.

Factor numerator and denominator:

$f(x) = \frac{x^2-1}{x^3-1} = \frac{(x-1)(x+1)}{(x-1)(x^2+x+1)} = \frac{x+1}{x^2+x+1}, \quad x\neq 1.$

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$f$ is discontinuous where its denominator is zero; factor $x^3+3x^2-x-3$ to find those points.

Points of discontinuity of a rational function occur where the denominator vanishes — a CBSE/NCERT Class 12 continuity topic.

Factor the denominator by grouping:

$x^3 + 3x^2 - x - 3 = x^2(x+3) - (x+3) = (x+3)(x^2-1) = (x+3)(x-1)(x+1).$

The function $f(x)=\dfrac{x^2+4x+3}{(x+3)(x-1)(x+1)}$ is undefined (hence discontinuous) wherever the denominator is zero:

$x = 1,\ -1,\ -3.$

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f(x) = |2x-3|/(2x-3) is the sign function of (2x-3); it is undefined at x = 3/2 and jumps from -1 to +1 there, so it is discontinuous at x = 3/2.

For $x > \frac{3}{2}$, $2x-3>0$, so $|2x-3|=2x-3$ and $f(x) = \frac{2x-3}{2x-3} = 1$.

For $x < \frac{3}{2}$, $2x-3<0$, so $|2x-3|=-(2x-3)$ and $f(x) = \frac{-(2x-3)}{2x-3} = -1$.

At $x=\frac{3}{2}$ itself, the denominator $2x-3=0$, so $f(x)$ is not even defined there.

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This function is continuous at exactly the one point where its two branches meet, $x=\dfrac12$.

$f(x)=x$ for irrational $x$ and $f(x)=1-x$ for rational $x$. Near any point $x=c$, both rationals and irrationals arbitrarily close to $c$ exist, so for $f$ to be continuous at $c$ we need the two branch formulas to agree there: $c=1-c\Rightarrow c=\dfrac12$.

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Use the standard limit $\displaystyle\lim_{x\to0}(1+ax)^{1/x}=e^a$.

For continuity at $x=0$: $f(0) = \displaystyle\lim_{x\to0}(1+2x)^{1/x}$

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$\tan x$ is continuous everywhere it is defined; it is undefined exactly at odd multiples of $\pi/2$, so among the given intervals only the open interval $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ avoids those points entirely.

$\tan x = \frac{\sin x}{\cos x}$ is undefined wherever $\cos x = 0$, i.e. at $x = \pm\pi/2, \pm3\pi/2,\ldots$

1. $\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ — closed interval, includes the undefined endpoints $\pm\pi/2$. Not valid.
2. $(-\infty,\infty)$ — includes infinitely many undefined points. Not valid. …