Questions
Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$
Direct substitution of $x=3$ gives the form $\frac{0}{0}$.
Factorize the numerator: $x^2 - 9 = (x-3)(x+3)$.
The limit becomes: $\lim_{x \to 3} \frac{(x-3)(x+3)}{x-3}$.
Cancel the common factor $(x-3)$ (since $x \to 3, x \neq 3$): $\lim_{x \to 3} (x+3)$.
Substitute $x=3$: $3 + 3 = 6$.
$6$
Evaluate $\lim_{x \to 1/2} \frac{4x^2 - 1}{2x - 1}$
Direct substitution of $x=1/2$ gives $\frac{0}{0}$.
Factorize the numerator: $4x^2 - 1 = (2x)^2 - 1^2 = (2x-1)(2x+1)$.
Limit becomes: $\lim_{x \to 1/2} \frac{(2x-1)(2x+1)}{2x-1}$.
Cancel $(2x-1)$: $\lim_{x \to 1/2} (2x+1)$.
Substitute $x=1/2$: $2(1/2) + 1 = 1 + 1 = 2$.
$2$
Evaluate $\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$
Rationalize the numerator by multiplying and dividing by $\sqrt{x+h} + \sqrt{x}$.
$\lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}$.
Simplify numerator: $(x+h) - x = h$.
Limit becomes: $\lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$.
Cancel $h$: $\lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$.
Substitute $h=0$: $\frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$.
$\frac{1}{2\sqrt{x}}$
Evaluate $\lim_{x \to 0} \frac{(x+2)^{1/3} - 2^{1/3}}{x}$
Let $y = x + 2$. As $x \to 0$, $y \to 2$. Also $x = y - 2$.
Substitute in limit: $\lim_{y \to 2} \frac{y^{1/3} - 2^{1/3}}{y - 2}$.
Use standard formula $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$.
Here $n = 1/3$ and $a = 2$.
Result: $\frac{1}{3}(2)^{1/3 - 1} = \frac{1}{3}(2)^{-2/3} = \frac{1}{3 \cdot 2^{2/3}}$.
$\frac{1}{3 \cdot 2^{2/3}}$
Evaluate $\lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^2 - 1}$
Let $y = 1+x$. As $x \to 0$, $y \to 1$.
Limit becomes: $\lim_{y \to 1} \frac{y^6 - 1}{y^2 - 1}$.
Divide numerator and denominator by $y-1$: $\frac{\lim_{y \to 1} \frac{y^6 - 1^6}{y - 1}}{\lim_{y \to 1} \frac{y^2 - 1^2}{y - 1}}$.
Using formula $n a^{n-1}$: Numerator $= 6(1)^5 = 6$. Denominator $= 2(1)^1 = 2$.
Result: $6/2 = 3$.
$3$
Evaluate $\lim_{x \to a} \frac{(x+2)^{5/2} - (a+2)^{5/2}}{x - a}$
Let $y = x+2$ and $b = a+2$. As $x \to a$, $y \to b$.
Also $x - a = (y-2) - (b-2) = y - b$.
Limit becomes: $\lim_{y \to b} \frac{y^{5/2} - b^{5/2}}{y - b}$.
Using formula $n b^{n-1}$: $\frac{5}{2} b^{5/2 - 1} = \frac{5}{2} b^{3/2}$.
Substitute back $b = a+2$: $\frac{5}{2} (a+2)^{3/2}$.
$\frac{5}{2} (a+2)^{3/2}$
Evaluate $\lim_{x \to 1} \frac{x^4 - \sqrt{x}}{\sqrt{x} - 1}$
Let $x = t^2$. As $x \to 1$, $t \to 1$.
Substitute: $\lim_{t \to 1} \frac{(t^2)^4 - t}{t - 1} = \lim_{t \to 1} \frac{t^8 - t}{t - 1}$.
Factor out $t$: $\lim_{t \to 1} t \cdot \frac{t^7 - 1}{t - 1}$.
Apply limit: $1 \cdot (7(1)^{6}) = 7$.
$7$
Evaluate $\lim_{x \to 2} \frac{x^2 - 4}{\sqrt{3x - 2} - \sqrt{x + 2}}$
Rationalize the denominator by multiplying numerator and denominator by $\sqrt{3x - 2} + \sqrt{x + 2}$.
Denominator becomes: $(3x - 2) - (x + 2) = 2x - 4 = 2(x - 2)$.
Limit: $\lim_{x \to 2} \frac{(x-2)(x+2)(\sqrt{3x - 2} + \sqrt{x + 2})}{2(x - 2)}$.
Cancel $(x-2)$: $\lim_{x \to 2} \frac{(x+2)(\sqrt{3x - 2} + \sqrt{x + 2})}{2}$.
Substitute $x=2$: $\frac{(4)(\sqrt{4} + \sqrt{4})}{2} = \frac{4(2+2)}{2} = \frac{16}{2} = 8$.
$8$
Evaluate $\lim_{x \to 2} \frac{x^4 - 16}{x^3 - 8}$
Rewrite as: $\lim_{x \to 2} \frac{x^4 - 2^4}{x^3 - 2^3}$.
Divide numerator and denominator by $x-2$: $\frac{\lim \frac{x^4 - 2^4}{x - 2}}{\lim \frac{x^3 - 2^3}{x - 2}}$.
Apply formula $n a^{n-1}$: $\frac{4(2)^3}{3(2)^2}$.
Simplify: $\frac{4 \cdot 8}{3 \cdot 4} = \frac{8}{3}$.
$8/3$
Evaluate $\lim_{x \to 1} \frac{x^7 - 2x^5 + 1}{x^3 - 3x^2 + 2}$
At $x=1$, form is $0/0$.
Rewrite Numerator: $x^7 - x^5 - x^5 + 1 = x^5(x^2-1) - (x^5-1)$.
Rewrite Denominator: $x^3 - 1 - 3x^2 + 3 = (x^3-1) - 3(x^2-1)$.
Divide both by $x-1$ and apply limits:
Num: $\lim_{x \to 1} [x^5(x+1) - (x^4+x^3+x^2+x+1)] = 1(2) - 5 = -3$.
Denom: $\lim_{x \to 1} [(x^2+x+1) - 3(x+1)] = 3 - 6 = -3$.
Result: $\frac{-3}{-3} = 1$.
$1$
Evaluate $\lim_{x \to 0} \frac{\sqrt{1+x^3} - \sqrt{1-x^3}}{x^2}$
Rationalize numerator: Multiply by $\sqrt{1+x^3} + \sqrt{1-x^3}$.
Num becomes: $(1+x^3) - (1-x^3) = 2x^3$.
Limit: $\lim_{x \to 0} \frac{2x^3}{x^2(\sqrt{1+x^3} + \sqrt{1-x^3})} = \lim_{x \to 0} \frac{2x}{\sqrt{1+x^3} + \sqrt{1-x^3}}$.
Substitute $x=0$: $\frac{0}{\sqrt{1} + \sqrt{1}} = 0$.
$0$
Evaluate $\lim_{x \to -3} \frac{x^3 + 27}{x^5 + 243}$
Rewrite as $\lim_{x \to -3} \frac{x^3 - (-3)^3}{x^5 - (-3)^5}$.
Divide num and den by $x - (-3)$ and use formula $n a^{n-1}$.
Result: $\frac{3(-3)^2}{5(-3)^4} = \frac{3 \cdot 9}{5 \cdot 81} = \frac{27}{405} = \frac{1}{15}$.
$1/15$
Evaluate $\lim_{x \to 1/2} \left( \frac{8x^3 - 1}{2x - 1} - \frac{4x^2 - 1}{2x - 1} \right)$
Combine terms: $\lim_{x \to 1/2} \frac{8x^3 - 4x^2 - 1 + 1}{2x - 1} = \lim_{x \to 1/2} \frac{8x^3 - 4x^2}{2x - 1}$.
Factor numerator: $4x^2(2x - 1)$.
Cancel $(2x-1)$: $\lim_{x \to 1/2} 4x^2$.
Substitute $x=1/2$: $4(1/4) = 1$.
$1$
Find $n$, if $\lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = 80, n \in N$
Using formula, LHS $= n 2^{n-1}$.
Given $n 2^{n-1} = 80$.
Factorize 80: $80 = 5 \times 16 = 5 \times 2^4$.
Comparing $n 2^{n-1}$ with $5 \cdot 2^4$, we get $n = 5$.
$5$
Evaluate $\lim_{x \to 0} \frac{\sin 3x}{\sin 7x}$
Rewrite as: $\lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3x \cdot \frac{1}{\frac{\sin 7x}{7x} \cdot 7x}$.
Using $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
Limit becomes: $1 \cdot \frac{3x}{7x} \cdot 1 = \frac{3}{7}$.
$3/7$
Evaluate $\lim_{x \to 0} \frac{\sin^2 2x}{\sin^2 4x}$
Rewrite as: $\left( \frac{\sin 2x}{2x} \cdot 2x \right)^2 \div \left( \frac{\sin 4x}{4x} \cdot 4x \right)^2$.
Limit becomes: $\frac{(1 \cdot 2x)^2}{(1 \cdot 4x)^2} = \frac{4x^2}{16x^2} = \frac{1}{4}$.
$1/4$
Evaluate $\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}$
Use identity $1 - \cos 2x = 2\sin^2 x$.
Limit: $\lim_{x \to 0} \frac{2\sin^2 x}{x^2} = 2 \left( \lim_{x \to 0} \frac{\sin x}{x} \right)^2$.
Result: $2(1)^2 = 2$.
$2$
Evaluate $\lim_{x \to 0} \frac{2\sin x - \sin 2x}{x^3}$
Use $\sin 2x = 2\sin x \cos x$.
Numerator: $2\sin x - 2\sin x \cos x = 2\sin x(1 - \cos x)$.
Use $1 - \cos x = 2\sin^2(x/2)$. Num $= 4\sin x \sin^2(x/2)$.
Limit: $\lim_{x \to 0} \frac{4\sin x}{x} \cdot \frac{\sin^2(x/2)}{(x/2)^2 \cdot 4} \cdot \frac{1}{x^2} \cdot x^2$ (Adjusting terms).
Rewrite: $4 \cdot \frac{\sin x}{x} \cdot \left(\frac{\sin(x/2)}{x/2}\right)^2 \cdot \frac{1}{4}$.
Result: $4(1)(1)^2(1/4) = 1$.
$1$
Evaluate $\lim_{x \to 0} \frac{1 - \cos mx}{1 - \cos nx}$
Use $1 - \cos \theta = 2\sin^2(\theta/2)$.
Limit: $\frac{2\sin^2(mx/2)}{2\sin^2(nx/2)} = \frac{\sin^2(mx/2)}{(mx/2)^2} \cdot (mx/2)^2 \div \left[ \frac{\sin^2(nx/2)}{(nx/2)^2} \cdot (nx/2)^2 \right]$.
As $x \to 0$, limit terms go to 1.
Remaining: $\frac{(mx/2)^2}{(nx/2)^2} = \frac{m^2 x^2 / 4}{n^2 x^2 / 4} = \frac{m^2}{n^2}$.
$m^2/n^2$
Evaluate $\lim_{x \to \pi/3} \frac{\sqrt{1 - \cos 6x}}{\sqrt{2}(\frac{\pi}{3} - x)}$
Let $x = \frac{\pi}{3} - h$. As $x \to \frac{\pi}{3}$, $h \to 0$.
Denominator: $\sqrt{2}h$.
Numerator: $\sqrt{1 - \cos 6(\frac{\pi}{3} - h)} = \sqrt{1 - \cos(2\pi - 6h)} = \sqrt{1 - \cos 6h}$.
$\sqrt{2\sin^2 3h} = \sqrt{2}|\sin 3h|$.
Limit: $\lim_{h \to 0} \frac{\sqrt{2}|\sin 3h|}{\sqrt{2}h} = \lim_{h \to 0} \frac{|\sin 3h|}{h}$.
Assuming $x < \pi/3$ (LHL) or principal value context, limit is 3. (Note: Strictly LHL=3, RHL=-3).
$3$
Evaluate $\lim_{x \to \pi/4} \frac{\sin x - \cos x}{x - \frac{\pi}{4}}$
Factor out $\sqrt{2}$: $\sqrt{2}(\frac{1}{\sqrt{2}}\sin x - \frac{1}{\sqrt{2}}\cos x) = \sqrt{2}(\sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4})$.
Numerator becomes $\sqrt{2}\sin(x - \pi/4)$.
Let $y = x - \pi/4$. Limit: $\lim_{y \to 0} \frac{\sqrt{2}\sin y}{y}$.
Result: $\sqrt{2}(1) = \sqrt{2}$.
$\sqrt{2}$
Evaluate $\lim_{x \to \pi/6} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{\pi}{6}}$
Factor out 2: $2(\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x)$.
Write as $2(\sin x \cos \frac{\pi}{6} - \cos x \sin \frac{\pi}{6}) = 2\sin(x - \pi/6)$.
Let $y = x - \pi/6$. Limit: $\lim_{y \to 0} \frac{2\sin y}{y} = 2$.
$2$
Evaluate $\lim_{x \to 0} \frac{\sin 2x + 3x}{2x + \tan 3x}$
Divide numerator and denominator by $x$.
$\frac{\frac{\sin 2x}{x} + 3}{2 + \frac{\tan 3x}{x}} = \frac{2\frac{\sin 2x}{2x} + 3}{2 + 3\frac{\tan 3x}{3x}}$.
Apply limits: $\frac{2(1) + 3}{2 + 3(1)} = \frac{5}{5} = 1$.
$1$
Evaluate $\lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}}$
Use $\sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}$.
Numerator: $2\cos\frac{x+a}{2}\sin\frac{x-a}{2}$.
Multiply and divide by $\sqrt{x} + \sqrt{a}$. Denom becomes $x-a$.
Limit: $\lim_{x \to a} \frac{\sin(\frac{x-a}{2})}{\frac{x-a}{2}} \cdot \frac{1}{2} \cdot 2\cos\frac{x+a}{2} \cdot (\sqrt{x} + \sqrt{a})$.
Result: $1 \cdot \cos a \cdot 2\sqrt{a} = 2\sqrt{a}\cos a$.
$2\sqrt{a}\cos a$
Evaluate $\lim_{x \to \pi/6} \frac{\cot^2 x - 3}{\csc x - 2}$
Use $\cot^2 x = \csc^2 x - 1$.
Numerator: $\csc^2 x - 1 - 3 = \csc^2 x - 4 = (\csc x - 2)(\csc x + 2)$.
Cancel $\csc x - 2$. Limit: $\lim_{x \to \pi/6} (\csc x + 2)$.
Substitute $x = \pi/6$: $\csc(\pi/6) + 2 = 2 + 2 = 4$.
$4$
Evaluate $\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^2 x}$
Rationalize numerator: Multiply by $\sqrt{2} + \sqrt{1 + \cos x}$.
Num: $2 - (1 + \cos x) = 1 - \cos x$.
Denom: $\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x}) = (1 - \cos^2 x)(\dots) = (1-\cos x)(1+\cos x)(\dots)$.
Cancel $1 - \cos x$. Limit: $\frac{1}{(1+\cos 0)(\sqrt{2} + \sqrt{1+\cos 0})}$.
Result: $\frac{1}{2(\sqrt{2} + \sqrt{2})} = \frac{1}{4\sqrt{2}}$.
$\frac{1}{4\sqrt{2}}$
Evaluate $\lim_{x \to 0} \frac{\sin x - 2\sin 3x + \sin 5x}{x}$
Split the limit: $\frac{\sin x}{x} - 2\frac{\sin 3x}{x} + \frac{\sin 5x}{x}$.
Apply standard limits: $1 - 2(3) + 5$.
Result: $1 - 6 + 5 = 0$.
$0$
If $\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}$, then find the value of $k$.
LHS: Using $n a^{n-1}$, limit is $4(1)^3 = 4$.
RHS: $\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \frac{3k^2}{2k} = \frac{3}{2}k$.
Equate: $4 = \frac{3}{2}k$.
Solve for $k$: $k = \frac{8}{3}$.
$8/3$
Differentiate w.r.t $x$: $x^4 + x^3 + x^2 + 1$
Let $y = x^4 + x^3 + x^2 + 1$.
Differentiate term by term using $\frac{d}{dx}(x^n) = nx^{n-1}$.
$\frac{dy}{dx} = 4x^3 + 3x^2 + 2x + 0$.
$4x^3 + 3x^2 + 2x$
Differentiate w.r.t $x$: $x + \frac{1}{x}$
Rewrite as $y = x + x^{-1}$.
$\frac{dy}{dx} = 1 + (-1)x^{-2}$.
Result: $1 - \frac{1}{x^2}$.
$1 - \frac{1}{x^2}$
Differentiate w.r.t $x$: $(3x + 5)(1 + \tan x)$
Use Product Rule: $u v' + v u'$.
Let $u = 3x+5, v = 1+\tan x$. Then $u' = 3, v' = \sec^2 x$.
$\frac{dy}{dx} = (3x+5)(\sec^2 x) + (1+\tan x)(3)$.
$3(1+\tan x) + (3x+5)\sec^2 x$
Differentiate w.r.t $x$: $(\sec x - 1)(\sec x + 1)$
Simplify first: $y = \sec^2 x - 1 = \tan^2 x$.
Use Chain Rule: $\frac{d}{dx}(\tan x)^2 = 2\tan x \cdot \frac{d}{dx}(\tan x)$.
Result: $2\tan x \sec^2 x$.
$2\tan x \sec^2 x$
Differentiate w.r.t $x$: $\frac{3x + 4}{5x^2 - 7x + 9}$
Use Quotient Rule: $\frac{v u' - u v'}{v^2}$.
$u = 3x+4, u' = 3$. $v = 5x^2-7x+9, v' = 10x-7$.
Num: $(5x^2-7x+9)(3) - (3x+4)(10x-7)$.
Expand Num: $15x^2 - 21x + 27 - (30x^2 - 21x + 40x - 28)$.
Simplify: $15x^2 - 21x + 27 - 30x^2 - 19x + 28 = -15x^2 - 40x + 55$.
$\frac{55 - 40x - 15x^2}{(5x^2 - 7x + 9)^2}$
Differentiate w.r.t $x$: $\frac{x^5 - \cos x}{\sin x}$
Use Quotient Rule with $u = x^5 - \cos x$ and $v = \sin x$.
$u' = 5x^4 + \sin x$, $v' = \cos x$.
$\frac{dy}{dx} = \frac{\sin x(5x^4 + \sin x) - (x^5 - \cos x)\cos x}{\sin^2 x}$.
Simplify numerator: $5x^4\sin x + \sin^2 x - x^5\cos x + \cos^2 x$.
Since $\sin^2 x + \cos^2 x = 1$, Num $= 5x^4\sin x - x^5\cos x + 1$.
$\frac{5x^4\sin x - x^5\cos x + 1}{\sin^2 x}$
Differentiate w.r.t $x$: $\frac{x^2 \cos(\frac{\pi}{4})}{\sin x}$
Constant $k = \cos(\pi/4) = 1/\sqrt{2}$. Function is $k \cdot \frac{x^2}{\sin x}$.
Differentiate $\frac{x^2}{\sin x}$ using Quotient Rule.
$\frac{\sin x(2x) - x^2(\cos x)}{\sin^2 x}$.
$\frac{1}{\sqrt{2}} \frac{2x\sin x - x^2\cos x}{\sin^2 x}$
Differentiate w.r.t $x$: $(ax^2 + \cot x)(p + q \cos x)$
Use Product Rule: $u v' + v u'$.
$u = ax^2 + \cot x, u' = 2ax - \csc^2 x$.
$v = p + q\cos x, v' = -q\sin x$.
$\frac{dy}{dx} = (ax^2 + \cot x)(-q\sin x) + (p + q\cos x)(2ax - \csc^2 x)$.
$-(ax^2 + \cot x)(q\sin x) + (p + q\cos x)(2ax - \csc^2 x)$
Differentiate w.r.t $x$: $\frac{a + b \sin x}{c + d \cos x}$
Use Quotient Rule. $u = a+b\sin x, v = c+d\cos x$.
$u' = b\cos x, v' = -d\sin x$.
Num: $(c+d\cos x)(b\cos x) - (a+b\sin x)(-d\sin x)$.
Expand: $bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x$.
Group terms: $bc\cos x + ad\sin x + bd(\cos^2 x + \sin^2 x) = bc\cos x + ad\sin x + bd$.
$\frac{bd + bc\cos x + ad\sin x}{(c + d\cos x)^2}$
Differentiate w.r.t $x$: $(\sin x + \cos x)^2$
Simplify first: $\sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x$.
Differentiate $1 + \sin 2x$.
Derivative of 1 is 0. Derivative of $\sin 2x$ is $2\cos 2x$.
$2\cos 2x$
Differentiate w.r.t $x$: $(2x - 7)^2 (3x + 5)^3$
Use Product Rule and Chain Rule.
$\frac{dy}{dx} = (2x-7)^2 \cdot 3(3x+5)^2(3) + (3x+5)^3 \cdot 2(2x-7)(2)$.
$= 9(2x-7)^2(3x+5)^2 + 4(2x-7)(3x+5)^3$.
Factor out $(2x-7)(3x+5)^2$: $(2x-7)(3x+5)^2 [9(2x-7) + 4(3x+5)]$.
Simplify bracket: $18x - 63 + 12x + 20 = 30x - 43$.
$(2x-7)(3x+5)^2(30x-43)$
Differentiate w.r.t $x$: $x^2 \sin x + \cos 2x$
Differentiate term by term.
For $x^2 \sin x$, use Product Rule: $x^2(\cos x) + \sin x(2x)$.
For $\cos 2x$, derivative is $-2\sin 2x$.
$x^2\cos x + 2x\sin x - 2\sin 2x$
Differentiate w.r.t $x$: $\sin^3 x \cos^3 x$
Rewrite as $(\sin x \cos x)^3 = (\frac{1}{2} \sin 2x)^3 = \frac{1}{8} \sin^3 2x$.
Differentiate using Chain Rule: $\frac{1}{8} \cdot 3\sin^2 2x \cdot \cos 2x \cdot 2$.
Simplify: $\frac{6}{8} \sin^2 2x \cos 2x = \frac{3}{4} \sin^2 2x \cos 2x$.
$\frac{3}{4} \sin^2 2x \cos 2x$
Differentiate w.r.t $x$: $\frac{1}{ax^2 + bx + c}$
Rewrite as $(ax^2 + bx + c)^{-1}$.
Use Chain Rule: $-1(ax^2 + bx + c)^{-2} \cdot \frac{d}{dx}(ax^2 + bx + c)$.
Inner derivative is $2ax + b$.
Result: $\frac{-(2ax + b)}{(ax^2 + bx + c)^2}$.
$\frac{-(2ax + b)}{(ax^2 + bx + c)^2}$