Q.Find the limits:

Standard result: $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$.

This is a well-known standard limit.

Factor $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + \cdots + a^{n-1})$.

So $\dfrac{x^n-a^n}{x-a} = x^{n-1} + x^{n-2}a + \cdots + a^{n-1}$ ($n$ terms).

As $x \to a$, each term $\to a^{n-1}$, and there are $n$ terms, giving $na^{n-1}$.

$\dfrac{\sin ax}{bx} = \dfrac{a}{b}\cdot\dfrac{\sin ax}{ax} \to \dfrac{a}{b}$.

Rewrite to use the standard limit $\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1$:

$\dfrac{\sin ax}{bx} = \dfrac{a}{b} \cdot \dfrac{\sin ax}{ax}$.

As $x \to 0$, $ax \to 0$, so $\dfrac{\sin ax}{ax} \to 1$.

Therefore the limit $= \dfrac{a}{b}\cdot 1 = \dfrac{a}{b}$.

For $x$ just above $3$, $[x]=3$, so $\frac{x}{[x]}\to\frac{3}{3}=1$.

Approaching from the right, $x \to 3^{+}$ means $3 < x < 4$, so the greatest integer $[x] = 3$.

Thus $\dfrac{x}{[x]} = \dfrac{x}{3}$, and as $x \to 3^{+}$ this $\to \dfrac{3}{3} = 1$.

The limit is $0$.

Using the identity $1-\cos2x=2\sin^{2}x$,

$\frac{1-\cos2x}{x}=\frac{2\sin^{2}x}{x}=2\sin x\cdot\frac{\sin x}{x}.$

As $x\to0$, $\frac{\sin x}{x}\to1$ and $\sin x\to0$, so

$\lim_{x\to0}2\sin x\cdot\frac{\sin x}{x}=2\cdot0\cdot1=0.$

The left and right limits are $-1$ and $+1$, so the limit does not exist.

For $x>0$, $|x|=x$, so $\frac{|x|}{x}=1$; thus $\lim_{x\to0^{+}}\frac{|x|}{x}=1$.

For $x<0$, $|x|=-x$, so $\frac{|x|}{x}=-1$; thus $\lim_{x\to0^{-}}\frac{|x|}{x}=-1$.

Since the two one-sided limits are unequal, $\lim_{x\to0}\frac{|x|}{x}$ does not exist.

The limit is $0$.

Rewrite

$\sec x-\tan x=\frac{1}{\cos x}-\frac{\sin x}{\cos x}=\frac{1-\sin x}{\cos x}.$

This is $\frac{0}{0}$ at $x=\frac{\pi}{2}$. Multiply numerator and denominator by $(1+\sin x)$:

$\frac{(1-\sin x)(1+\sin x)}{\cos x(1+\sin x)}=\frac{1-\sin^{2}x}{\cos x(1+\sin x)}=\frac{\cos^{2}x}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}.$

As $x\to\frac{\pi}{2}$, $\cos x\to0$ and $1+\sin x\to2$, so the limit is $\frac{0}{2}=0$.

The limit of a constant function is that constant itself.

Here $f(x)=2024$ for all real $x$ — it is a constant function, so its graph is a horizontal line at height $2024$. As $x\to 0$ (or as $x$ approaches any value), $f(x)$ stays exactly at $2024$; there is no variation to consider. Formally, for the constant function $f(x)=c$, $\lim\limits_{x\to a} f(x) = c$ for every $a$.