JEE Mains: Limits, Continuity, Different

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1. Limits 1.1 Introduction to Limits A function $f(x)$ has a limit $L$ at $x=a$ if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$. Left Hand Limit (LHL): $\lim_{x \to a^-} f(x)$ Right Hand Limit (RHL): $\lim_{x \to a^+} f(x)$ 1.2 Methods for Evaluating Limits Direct Substitution: If $f(a)$ is defined and finite. Factorization: For indeterminate forms like $\frac{0}{0}$. Factorize numerator and denominator to cancel common factors. Rationalization: For expressions involving square roots, multiply by the conjugate. L'Hôpital's Rule: For indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$. If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$, provided the latter limit exists. Standard Limits: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ $\lim_{x \to 0} \frac{\tan x}{x} = 1$ $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$ $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \quad (a>0)$ $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$ $\lim_{x \to \infty} (1 + \frac{1}{x})^x = e$ $\lim_{x \to 0} (1 + x)^{1/x} = e$ $\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1$ $\lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1$ Limits of the form $1^\infty$: If $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \infty$, then $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} [f(x)-1]g(x)}$ 1.3 Sandwich Theorem (Squeeze Play Theorem) If $g(x) \le f(x) \le h(x)$ for all $x$ in an open interval containing $c$ (except possibly at $c$ itself), and if $\lim_{x \to c} g(x) = L$ and $\lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$. 2. Continuity 2.1 Definition of Continuity at a Point A function $f(x)$ is continuous at $x=a$ if: $f(a)$ exists (is defined). $\lim_{x \to a} f(x)$ exists. $\lim_{x \to a} f(x) = f(a)$. Equivalently, LHL = RHL = $f(a)$. 2.2 Types of Discontinuity Removable Discontinuity: $\lim_{x \to a} f(x)$ exists but is not equal to $f(a)$ (if $f(a)$ is defined). Or $f(a)$ is undefined. Can be removed by redefining $f(a)$. E.g., $f(x) = \frac{\sin x}{x}$ at $x=0$. Non-Removable Discontinuity: Jump Discontinuity: LHL $\ne$ RHL. E.g., $f(x) = [x]$ (greatest integer function). Infinite Discontinuity: One or both limits are $\pm \infty$. E.g., $f(x) = \frac{1}{x-a}$. Oscillatory Discontinuity: E.g., $f(x) = \sin(\frac{1}{x})$ at $x=0$. 2.3 Continuity in an Interval Open Interval $(a,b)$: $f(x)$ is continuous at every point in $(a,b)$. Closed Interval $[a,b)$: $f(x)$ is continuous in $(a,b)$, $\lim_{x \to a^+} f(x) = f(a)$. Closed Interval $(a,b]$: $f(x)$ is continuous in $(a,b)$, $\lim_{x \to b^-} f(x) = f(b)$. Closed Interval $[a,b]$: $f(x)$ is continuous in $(a,b)$, $\lim_{x \to a^+} f(x) = f(a)$, and $\lim_{x \to b^-} f(x) = f(b)$. 2.4 Properties of Continuous Functions If $f$ and $g$ are continuous at $x=a$, then $f \pm g$, $f \cdot g$, $f/g$ ($g(a) \ne 0$), $c \cdot f$ are also continuous at $x=a$. Composition of continuous functions is continuous: If $g$ is continuous at $x=a$ and $f$ is continuous at $g(a)$, then $f(g(x))$ is continuous at $x=a$. All polynomial functions, exponential functions $a^x$, logarithmic functions $\log_a x$, trigonometric functions are continuous in their domains. 2.5 Intermediate Value Theorem (IVT) If $f(x)$ is continuous on $[a,b]$ and $f(a) \ne f(b)$, then for every value $k$ between $f(a)$ and $f(b)$, there exists at least one $c \in (a,b)$ such that $f(c) = k$. Corollary: If $f(a)$ and $f(b)$ have opposite signs, then there is at least one root of $f(x)=0$ in $(a,b)$. 3. Differentiability 3.1 Definition of Differentiability at a Point A function $f(x)$ is differentiable at $x=a$ if $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists and is finite. This limit is called the derivative of $f(x)$ at $x=a$, denoted by $f'(a)$. Left Hand Derivative (LHD): $f'(a^-) = \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$ Right Hand Derivative (RHD): $f'(a^+) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$ For $f(x)$ to be differentiable at $x=a$, LHD = RHD = finite value. 3.2 Relation between Continuity and Differentiability If a function is differentiable at a point, it must be continuous at that point. Differentiability $\implies$ Continuity The converse is NOT necessarily true. A function can be continuous at a point but not differentiable at that point. Example: $f(x) = |x|$ is continuous at $x=0$ but not differentiable at $x=0$. LHD at $x=0$: $\lim_{h \to 0^-} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$ RHD at $x=0$: $\lim_{h \to 0^+} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$ Since LHD $\ne$ RHD, $f(x)=|x|$ is not differentiable at $x=0$. 3.3 Points where Differentiability Fails Sharp corners or cusps: E.g., $f(x)=|x|$ at $x=0$. Vertical tangent: E.g., $f(x)=x^{1/3}$ at $x=0$. $f'(x) = \frac{1}{3}x^{-2/3}$, $f'(0)$ is undefined. Discontinuity: If a function is discontinuous at a point, it cannot be differentiable at that point. 3.4 Differentiability in an Interval A function is differentiable in an open interval $(a,b)$ if it is differentiable at every point in $(a,b)$. A function is differentiable in a closed interval $[a,b]$ if it is differentiable in $(a,b)$, and $\lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h}$ exists, and $\lim_{h \to 0^-} \frac{f(b+h)-f(b)}{h}$ exists. 3.5 Properties of Differentiable Functions If $f$ and $g$ are differentiable at $x=a$, then $f \pm g$, $f \cdot g$, $f/g$ ($g(a) \ne 0$), $c \cdot f$ are also differentiable at $x=a$. Chain Rule: If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. 3.6 Basic Differentiation Formulas $\frac{d}{dx}(c) = 0$ $\frac{d}{dx}(x^n) = nx^{n-1}$ $\frac{d}{dx}(e^x) = e^x$ $\frac{d}{dx}(a^x) = a^x \ln a$ $\frac{d}{dx}(\ln|x|) = \frac{1}{x}$ $\frac{d}{dx}(\log_a|x|) = \frac{1}{x \ln a}$ $\frac{d}{dx}(\sin x) = \cos x$ $\frac{d}{dx}(\cos x) = -\sin x$ $\frac{d}{dx}(\tan x) = \sec^2 x$ $\frac{d}{dx}(\cot x) = -\csc^2 x$ $\frac{d}{dx}(\sec x) = \sec x \tan x$ $\frac{d}{dx}(\csc x) = -\csc x \cot x$ $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$ $\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$ $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$