Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

For a function to remain continuous at a point, its left-hand limit, right-hand limit, and defined value at that point must all converge to the same result. At $x=0$, continuity demands that the limit from the negative direction, the limit from the positive direction, and the assigned value $q$ are equivalent.

Analyzing the left side, the expression $\frac{\sin((p+1)x)+\sin x}{x}$ can be split into the two standard limits $\frac{\sin((p+1)x)}{x}$ and $\frac{\sin x}{x}$. By utilizing the property $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, the first term evaluates to $(p+1)$ and the second to $1$, simplifying the combined left-hand limit to $p+2$.

Evaluating the right side, the expression $\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3/2}}$ simplifies by factoring $\sqrt{x}$ out of the numerator and denominator to produce $\frac{\sqrt{1+x}-1}{x}$. Rationalizing this fraction by multiplying the numerator and denominator by the conjugate $\sqrt{1+x}+1$ results in $\frac{(1+x)-1}{x(\sqrt{1+x}+1)}$, which reduces to $\frac{1}{\sqrt{1+x}+1}$. As $x$ approaches $0$, this expression evaluates to $\frac{1}{2}$. Equating the left-hand limit $p+2$ with the right-hand limit $\frac{1}{2}$ determines that $p=-\frac{3}{2}$, while setting $q$ equal to these values confirms $q=\frac{1}{2}$.

To determine if a limit exists, the values approaching from both the left and right sides must converge to the same result. The numerator $\sqrt{1-\cos\{2(x-2)\}}$ simplifies using the trigonometric identity $1-\cos(2\theta) = 2\sin^2(\theta)$ to $\sqrt{2\sin^2(x-2)}$, which is equivalent to $\sqrt{2}|\sin(x-2)|$, transforming the expression into $\lim_{x\to 2} \frac{\sqrt{2}|\sin(x-2)|}{x-2}$. When evaluating the left-hand limit as $x$ approaches $2^{-}$, the quantity $(x-2)$ is negative, causing the absolute value to introduce a negative sign that results in $\lim_{x\to 2^{-}} \frac{-\sqrt{2}\sin(x-2)}{x-2} = -\sqrt{2}$. Conversely, the right-hand limit as $x$ approaches $2^{+}$ involves a positive $(x-2)$, which keeps the expression positive and yields $\lim_{x\to 2^{+}} \frac{\sqrt{2}\sin(x-2)}{x-2} = \sqrt{2}$. Because the left-hand and right-hand limits are distinct, the limit does not exist.

The Squeeze Theorem determines the limit of this expression by bounding it between two functions that converge to the same value. Since $f(x)$ is a positive increasing function, we know that $f(x) \leq f(2x) \leq f(3x)$ for all $x$.

Dividing this inequality by the positive quantity $f(x)$ yields $1 \leq \frac{f(2x)}{f(x)} \leq \frac{f(3x)}{f(x)}$. Taking the limit as $x \to \infty$, the lower bound is $1$ and the upper bound approaches $1$, which forces the ratio $\frac{f(2x)}{f(x)}$ to also converge to $1$.

The composite function $(g \circ f)(x) = \sin(x|x|)$ is defined piecewise by considering the sign of $x$. For $x \ge 0$, the function simplifies to $\sin(x^2)$, and for $x < 0$, it simplifies to $\sin(-x^2)$, which is $-\sin(x^2)$. Differentiating these expressions yields the first derivative:
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$Since the limit of this derivative from both the left and right at$x=0$is$0$, the function is differentiable at the origin and its derivative is continuous, confirming that the first statement is true.

Determining the second derivative requires differentiating the first derivative piecewise:
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$Evaluating these expressions as$x$approaches$0$gives a left-hand limit of$-2$and a right-hand limit of$2$. Because these one-sided limits are not equal, the second derivative does not exist at the origin, meaning the second statement is false.

Differentiability at a point relies on the existence and equality of the left-hand and right-hand limits of the difference quotient. At $x=1$, calculating the difference quotient involves comparing the behavior of $\sin\frac{1}{h}$ from the right and its negative counterpart from the left; because these limits do not result in the same value, the function fails to be differentiable at this point. In contrast, for $x=0$, the function follows the definition $f(x)=(x-1)\sin\frac{1}{x-1}$. Using the product rule to differentiate this expression, we find $f'(x) = \sin\frac{1}{x-1} - \frac{1}{x-1}\cos\frac{1}{x-1}$, which evaluates to the finite value $-\sin 1 + \cos 1$ at $x=0$, confirming the function is differentiable at this location.

The definition of $f(x)=\min\{x+1,|x|+1\}$ depends on comparing its two component terms. Since $|x| \ge x$ holds true for any real number, adding 1 to both sides confirms that $|x|+1 \ge x+1$. Because $x+1$ is always the smaller of the two terms, the function simplifies strictly to $f(x)=x+1$. This resulting linear function is a polynomial, which ensures it is differentiable everywhere on its domain.

A function is continuous at a point when its defined value at that coordinate matches the limit of the function as the variable approaches that point. To find the required value for $f(0)$, we evaluate the limit as $x$ approaches $0$ for the expression $\frac{1}{x} - \frac{2}{e^{2x}-1}$. Finding a common denominator allows us to rewrite this as $\frac{e^{2x}-1-2x}{x(e^{2x}-1)}$.

As $x$ nears $0$, both the numerator and the denominator vanish, resulting in an indeterminate $\frac{0}{0}$ form. Differentiating the numerator and the denominator with respect to $x$ yields $\frac{2e^{2x}-2}{2xe^{2x}+e^{2x}-1}$. Because this expression also evaluates to $\frac{0}{0}$ at the limit, we apply the rule again to arrive at $\frac{4e^{2x}}{4e^{2x}+4xe^{2x}}$. Substituting $x=0$ into this final ratio results in $\frac{4}{4+0}$, which equals $1$. Defining $f(0)=1$ effectively bridges the gap at the origin, ensuring the function remains continuous.

Differentiability requires that a function be continuous and possess a defined, unique slope at every point in its domain. To analyze $f(x)=\frac{x}{1+|x|}$, we partition the function based on the definition of the absolute value: for $x < 0$, the expression simplifies to $f(x)=\frac{x}{1-x}$, and for $x \geq 0$, it becomes $f(x)=\frac{x}{1+x}$.

Since both individual pieces are continuous and differentiable in their respective regions, we only need to verify the point $x=0$. By examining the behavior of $f(x)$ as $x$ approaches $0$ from both sides, we find that the left-hand and right-hand derivatives are equal, confirming the function is smooth at the origin. Because there are no breaks or sharp turns across the entire real line, the function is differentiable for all $x$, spanning the set $(-\infty , \infty )$.

Since $\alpha$ and $\beta$ are roots of $ax^{2+}bx+c=0$, the quadratic expression can be factored as $a(x-\alpha)(x-\beta)$, allowing the limit to be written as $\lim_{x\to \alpha} \frac{1-\cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2}$. Utilizing the trigonometric identity $1-\cos(\theta) = 2\sin^2(\theta/2)$, the numerator becomes $2\sin^2\left(\frac{a(x-\alpha)(x-\beta)}{2}\right)$. Multiplying and dividing by $\frac{a^2(x-\alpha)^2(x-\beta)^2}{4}$ enables the application of the fundamental limit $\lim_{u\to 0} \frac{\sin u}{u} = 1$. As $x$ approaches $\alpha$, the trigonometric part tends to 1, and the expression simplifies to the limit of $\frac{2}{4} \cdot a^2 (x-\beta)^2$, which evaluates to $\frac{a^2(\alpha-\beta)^2}{2}$.

The derivative of a function at $x=1$ is defined by the expression $f'(1) = \lim_{h\to 0} \frac{f(1+h) - f(1)}{h}$. Because the provided limit $\lim_{h\to 0} \frac{f(1+h)}{h} = 5$ converges to a finite value, the numerator $f(1+h)$ must approach 0 as $h$ approaches 0, which establishes that $f(1) = 0$ given the function's continuity at that point. Substituting $f(1) = 0$ into the derivative definition shows that the expression is equivalent to the given limit, which confirms the final value is 5.

The fundamental behavior of a differentiable function is revealed by its difference quotient, which we isolate by substituting $y=x+h$ into the inequality $|f(x)-f(y)| \leq (x-y)^2$ to get $|f(x+h)-f(x)| \leq h^2$. Dividing both sides by $|h|$ yields $\left| \frac{f(x+h)-f(x)}{h} \right| \leq |h|$, and taking the limit as $h \to 0$ shows that the derivative $f'(x)$ must be zero. Since the derivative is zero everywhere, the function is constant, and the condition $f(0)=0$ forces $f(x)=0$ for all $x$, so $f(1)=0$.

Evaluating limits of the form $\lim_{x \to \infty } (1+f(x))^{g(x)}$ where the base approaches $1$ and the exponent approaches infinity relies on the exponential identity $e^{\lim_{x \to \infty } f(x)g(x)}$. In the given expression, $f(x)$ corresponds to $\frac{a}{x} + \frac{b}{x^2}$ and $g(x)$ is $2x$. The product of these two terms is $2x(\frac{a}{x} + \frac{b}{x^2})$, which simplifies to $2a + \frac{2b}{x}$. As $x$ increases without bound, the term containing $b$ approaches $0$, leaving only $2a$ in the exponent. Equating this limit to the target value $e^2$ results in $e^{2a} = e^2$, which determines that $a = 1$. Because the term $\frac{2b}{x}$ vanishes as $x$ tends to infinity, the result remains valid regardless of the specific value of $b$, meaning $b$ can be any real number.

The infinite structure of the exponent allows us to substitute the original variable $x$ back into the expression, simplifying the relationship to $x=e^{y+x}$. By applying the natural logarithm to both sides, we get the equation $\ln x = y + x$. Differentiating both sides with respect to $x$ yields the relationship $\frac{1}{x} = \frac{dy}{dx} + 1$. Isolating the derivative by subtracting 1 from both sides gives $\frac{dy}{dx} = \frac{1}{x} - 1$, which simplifies to the final form of $\frac{1-x}{x}$.

Continuity at a point requires the function's value to equal the limit as the variable approaches that point. Calculating $f\left(\frac{\pi}{4}\right)$ involves finding the limit $\lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{4x-\pi}$ by substituting $x = \frac{\pi}{4} + h$ where $h$ approaches zero. This substitution leads to the expression $\frac{1-\tan\left(\frac{\pi}{4}+h\right)}{4h}$, which simplifies using the tangent addition identity to $\frac{-2\tan h}{4h(1-\tan h)}$. Applying the fundamental limit $\lim_{h \to 0} \frac{\tan h}{h} = 1$ results in $\frac{-2}{4}$, which simplifies to $-\frac{1}{2}$.

Substituting $x=0$ into the limit $\lim_{x\to 0}\;\frac{\log(3+x)-\log(3-x)}{x}$ results in the indeterminate form $\frac{0}{0}$, which indicates that we should apply L'Hôpital's rule by differentiating both the numerator and the denominator with respect to $x$. Taking the derivative of the numerator gives $\frac{1}{3+x} - \frac{1}{3-x}(-1)$, which simplifies to $\frac{1}{3+x} + \frac{1}{3-x}$, while the derivative of the denominator $x$ is $1$. Evaluating this resulting expression as $x$ approaches $0$ yields $\frac{1}{3+0} + \frac{1}{3-0}$, which calculates to $\frac{1}{3} + \frac{1}{3}$, giving a final value of $k = \frac{2}{3}$.

The core of this problem lies in the trigonometric identity for $\tan(A-B)$, specifically noting that the fraction $\frac{1-\tan(x/2)}{1+\tan(x/2)}$ is equivalent to $\tan(\frac{\pi}{4}-\frac{x}{2})$. This simplification is powerful because it converts the complex numerator into a single tangent term, which interacts predictably with the $(1-\sin x)$ factor as $x$ approaches $\frac{\pi}{2}$.

Setting $x = \frac{\pi}{2} + y$ as $y$ approaches $0$ transforms the expression into a manageable limit involving $y$. The denominator $(\pi-2x)^3$ becomes $-8y^3$, while the numerator terms become $-\tan(\frac{y}{2})$ and $2\sin^2(\frac{y}{2})$, giving the expression:

$\lim_{y \to 0} \frac{-\tan\left(\frac{y}{2}\right) \cdot 2\sin^2\left(\frac{y}{2}\right)}{-8y^3}$

By rearranging this into standard forms $\frac{\tan(y/2)}{y}$ and $(\frac{\sin(y/2)}{y})^2$, and adjusting the coefficients by $\frac{1}{2}$ to match the internal arguments, the product simplifies to $\frac{1}{4} \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^2$, resulting in $\frac{1}{32}$.

The evaluation of the given limit depends on recognizing the indeterminate $0/0$ form that arises as $x$ approaches $a$, because $f(a) = g(a) = k$. Applying L'Hôpital's rule is the appropriate method here, which involves differentiating both the numerator and the denominator with respect to $x$. The numerator, which simplifies to $f(a)g(x) - g(a)f(x)$ due to the cancellation of the $-f(a)$ and $+f(a)$ terms, has a derivative of $f(a)g'(x) - g(a)f'(x)$. Dividing this by the derivative of the denominator, $g'(x) - f'(x)$, yields the limit $\lim_{x\to a} \frac{f(a)g'(x) - g(a)f'(x)}{g'(x) - f'(x)}$. Substituting $f(a) = k$ and $g(a) = k$ into this expression simplifies the numerator to $k(g'(x) - f'(x))$, allowing the common term to cancel out and leaving only the value $k$. Since the overall limit is given as 4, it follows that $k=4$.

A function is continuous at a point if its value equals the limit from both sides. For $x=0$, the right-hand limit is $\lim_{h \to 0} h e^{-2/h} = 0$, and the left-hand limit is $\lim_{h \to 0} (-h) e^0 = 0$, which correctly matches $f(0)=0$. Differentiability requires the right-hand and left-hand derivatives to be equal, but these values differ in this case. The right-hand derivative is calculated as $\lim_{h \to 0} \frac{h e^{-2/h}}{h} = 0$, while the left-hand derivative simplifies to $\lim_{h \to 0} \frac{-h}{-h} = 1$. Since the derivatives are not equal, the function is continuous everywhere but fails to be differentiable at $x=0$.

Continuity at a point requires the function's limit to equal its value at that point, meaning that as $x$ approaches any value $a$, the function must approach the same number regardless of whether it is being approached through rational or irrational numbers. For any non-zero $a$, the immediate neighborhood contains both types of values, causing the function to approach $a$ and $-a$ simultaneously, which prevents the existence of a limit. Therefore, the function is discontinuous everywhere except at $x=0$, where $f(0)=0$ and the limit successfully converges to that value.

Evaluating a limit at a point requires that the function approaches the same finite value from both the left and right sides. For the expression $\lim_{x\to 0}\frac{\log x^n - [x]}{[x]}$, the behavior of the greatest integer function $[x]$ creates an immediate obstacle, as it jumps from $-1$ to $0$ as $x$ crosses zero. Since the limit of this fundamental component does not exist, the entire ratio fails to approach a unique, well-defined value at this point.