Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

Rewriting the expression $\frac{\cot^3 x - \tan x}{\cos(x+\pi/4)}$ as $\frac{1-\tan^4 x}{\tan^3 x \cos(x+\pi/4)}$ simplifies the analysis as $x$ approaches $\pi/4$ because $\tan x$ approaches $1$. Factoring the numerator $1-\tan^4 x$ into $(1-\tan^2 x)(1+\tan^2 x)$ and expanding the denominator $\cos(x+\pi/4)$ as $\frac{\cos x - \sin x}{\sqrt{2}}$ helps reveal the shared algebraic terms. Since $\cos x - \sin x$ is equivalent to $\cos x(1-\tan x)$, one can cancel the common $(1-\tan x)$ factor from the numerator expression $(1-\tan x)(1+\tan x)(1+\tan^2 x)$. The simplified expression $\frac{\sqrt{2}(1+\tan x)(1+\tan^2 x)}{\tan^3 x \cos x}$ evaluates to $8$ when substituting $x = \pi/4$.

The differentiability of a function requires that the left-hand and right-hand derivatives be equal at any point where the definition potentially changes. For the function $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$, the only candidate for non-differentiability is $x=0$ due to the presence of absolute value terms. Analyzing the component $\sin |x| - |x|$, the right-hand derivative at $x=0$ is $\cos(0) - 1 = 0$, while the left-hand derivative is $-\cos(0) + 1 = 0$; since these match, this part of the function is differentiable at the origin. Furthermore, the term $2(x - \pi) \cos |x|$ simplifies to $2(x - \pi) \cos x$ because the cosine function is symmetric, and being a product of differentiable functions, it remains smooth everywhere. Consequently, the function has no points where it fails to be differentiable, meaning the set $K$ is equal to $\Phi$.

The differentiability of the composite function $g(x)=|f(x)|$ at a point $x=c$ where $f(c)=0$ relies entirely on the slope of the original function at that location. By examining the expression $g'(x) = \frac{|f(x)|}{f(x)} f'(x)$, we can see that if $f'(c)=0$, the slope approaches zero from both the left and right directions, effectively allowing the derivative to exist. This demonstrates that the function is differentiable when $f'(c)=0$.

A function is not differentiable at points where the graph displays sharp corners or where adjacent segments join with differing derivatives. The expression can be represented piecewise by comparing the magnitudes of $|x|$ and $x^2$, yielding $f(x) = 8+2x$ on $[-4, -2)$, $x^2$ on $[-2, -1]$, $|x|$ on $(-1, 1)$, $x^2$ on $[1, 2]$, and $8-2x$ on $(2, 4]$.

The boundaries reveal that the derivatives do not align across these transitions. At $x = -2$, the slope changes from 2 to -4, and at $x = -1$, it transitions from -2 to -1. The point $x = 0$ is a corner where the slope shifts from -1 to 1, while at $x = 1$, it moves from 1 to 2, and at $x = 2$, it drops from 4 to -2. These mismatches in the derivative at $x = -2, -1, 0, 1, 2$ define the set of points where the function is not differentiable.

Continuity at a specific point requires the left-hand limit, the right-hand limit, and the function value to be identical. Examining the function $f(x)$ over $[-1, 3]$, we analyze its behavior across the defined sub-intervals: $f(x)=-x-1$ for $-1 \le x < 0$, $f(x)=x$ for $0 \le x < 1$, $f(x)=2x$ for $1 \le x < 2$, $f(x)=x+2$ for $2 \le x < 3$, and $f(3)=6$.

At $x=0$, the left-hand limit $\lim_{x \to 0^-} f(x) = -1$ differs from the right-hand limit $\lim_{x \to 0^+} f(x) = 0$, establishing a jump discontinuity. At $x=1$, the left-hand limit $\lim_{x \to 1^-} f(x) = 1$ is distinct from the right-hand limit $\lim_{x \to 1^+} f(x) = 2$, identifying a second point of discontinuity. The function remains continuous at $x=2$ because $\lim_{x \to 2^-} f(x) = 4$ and $\lim_{x \to 2^+} f(x) = 4$, but at $x=3$, the left-hand limit $\lim_{x \to 3^-} f(x) = 5$ does not equal the function value $f(3)=6$, creating a third discontinuity. Because these are the only points where continuity conditions are violated, the function is discontinuous at only three points.

The function $f(x) = \min(\sin x, \cos x)$ is non-differentiable at points where the graphs of $\sin x$ and $\cos x$ intersect, provided their derivatives are unequal at those intersections. Setting $\sin x = \cos x$ yields $\tan x = 1$, which corresponds to $x = \frac{\pi}{4} + n\pi$ for any integer $n$. Restricting these points to the open interval $(-\pi, \pi)$ gives $x = \frac{\pi}{4}$ (for $n=0$) and $x = -\frac{3\pi}{4}$ (for $n=-1$). Since the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$, evaluating these at the intersection points shows the derivatives are unequal ($\frac{1}{\sqrt{2}} \neq -\frac{1}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}} \neq \frac{1}{\sqrt{2}}$), confirming non-differentiability at these two values. Thus, the set of non-differentiable points is $S = \{-\frac{3\pi}{4}, \frac{\pi}{4}\}$, which is a subset of $\{ -\frac{3\pi}{4} , -\frac{\pi}{4} , \frac{3\pi}{4} , \frac{\pi}{4} \}$.

Indeterminate forms of the type $0/0$ are best resolved by factoring expressions to cancel the terms that cause the division by zero. For the expression on the left, recognizing the numerator $x^4 - 1$ as a product of $(x-1)$ and the polynomial $(x^{3+}x^{2+}x+1)$ allows us to cancel the denominator and evaluate the limit as $4$ by substituting $x=1$.

For the expression on the right, factoring the difference of cubes $x^3 - k^3$ into $(x-k)(x^{2+}xk+k^2)$ and the difference of squares $x^2 - k^2$ into $(x-k)(x+k)$ allows the $(x-k)$ terms to cancel. Evaluating the resulting fraction $\frac{x^{2+}xk+k^2}{x+k}$ as $x$ approaches $k$ simplifies to $\frac{3k^2}{2k}$, which reduces to $\frac{3k}{2}$. Equating the two limits gives $\frac{3k}{2} = 4$, resulting in $k = \frac{8}{3}$.

The differentiability of a function at a specific point requires both continuity and the equality of the left-hand and right-hand derivatives. For a piecewise defined function, we must examine the behavior at every transition point to identify where the slope changes abruptly.

First, consider the term $f(|x|)$ for $x \in [-2, 2]$. Because the absolute value $|x|$ maps to the non-negative range $[0, 2]$, we consistently use the second piece of the definition, $f(|x|) = |x|^2 - 1$, which simplifies to $x^2 - 1$. Next, determine $|f(x)|$ by taking the absolute value of the original pieces, noting that $x^2 - 1$ changes sign at $x=1$. This yields $|f(x)| = 1$ for $x \in [-2, 0)$, $|f(x)| = 1 - x^2$ for $x \in [0, 1)$, and $|f(x)| = x^2 - 1$ for $x \in [1, 2]$.

Combining these results into $g(x) = |f(x)| + f(|x|)$ provides the piecewise definition:
$

$Checking the critical points for continuity, the function approaches$0$from both sides at$x=0$and at$x=1$, confirming the function is continuous everywhere. Finally, evaluating the derivative at$x=1$, the derivative of the left piece is$0$, while the derivative of the right piece,$2x^2 - 2$, is$4x$. At$x=1$, this right-hand derivative is$4$. Since the left-hand derivative$0$does not equal the right-hand derivative$4$, the function is not differentiable at this point.

When the variable $x$ approaches $0$ from the negative side, the greatest integer function $[x]$ is fixed at $-1$, while the absolute value function $|x|$ is defined as $-x$. Substituting these values into the expression $\frac{x([x]+|x|)\sin[x]}{|x|}$ yields the simplified form $\frac{x(-1-x)\sin(-1)}{-x}$. Canceling the term $-x$ from the denominator with the corresponding factors in the numerator simplifies the expression to $(1+x)\sin(-1)$, which is equivalent to $-(1+x)\sin 1$. Evaluating the limit as $x \to 0$ leads directly to the result $-\sin 1$.

The non-differentiable points of $f(x) = \max \{- |x| , - \sqrt{1-x^2}\}$ are identified by analyzing both the inherent non-differentiability of its components and the locations where they intersect. First, the component $-|x|$ is non-differentiable at $x = 0$, establishing this as the first point of concern. Second, the two functions intersect where $-|x| = -\sqrt{1-x^2}$, which simplifies to $|x| = \sqrt{1-x^2}$ and leads to $x^2 = 1 - x^2$, or $2x^2 = 1$, yielding the points $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$. Since the derivatives of the two components differ at these intersection points, the function $f(x)$ fails to be differentiable there as well. Consequently, there are exactly three points of non-differentiability within the interval $(-1, 1)$.

Limits involving a variable exponent that approach the indeterminate form $1^\infty$ can be evaluated by applying the identity $\lim_{x \to a} [g(x)]^{h(x)} = e^{\lim_{x \to a} h(x)[g(x)-1]}$. Applying this property to the expression $\lim_{x \to 0} \left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{\frac{1}{x}}$ involves subtracting one from the base, which simplifies the inner term to $\frac{f(3+x)-f(3) - f(2-x)+f(2)}{1+f(2-x)-f(2)}$. Multiplying this result by the exponent $\frac{1}{x}$ allows for the application of the limit as $x$ approaches zero.

The expression within the exponent then becomes $\frac{1}{x}\left(\frac{f(3+x)-f(3)}{1+f(2-x)-f(2)} - \frac{f(2-x)-f(2)}{1+f(2-x)-f(2)}\right)$. As $x$ tends to zero, the term $\frac{f(3+x)-f(3)}{x}$ approaches the derivative $f'(3)$. Similarly, the term $\frac{f(2-x)-f(2)}{x}$ can be rewritten as $-\frac{f(2-x)-f(2)}{-x}$, which approaches $-f'(2)$. Since the denominator $1+f(2-x)-f(2)$ approaches 1, the limit of the entire exponent simplifies directly to $f'(3) - (-f'(2))$, which is $f'(3)+f'(2)$. Given the condition $f'(3)+f'(2)=0$, the calculation results in $e^0$, giving a final value of 1.

A function is continuous at $x=0$ if $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$. Evaluating the left-hand limit, the expression becomes:
$\lim_{x \to 0^-} \frac{\sin(p+1)x + \sin x}{x} = \lim_{x \to 0^-} \left( \frac{\sin(p+1)x}{x} + \frac{\sin x}{x} \right) = (p+1) + 1 = p+2$
The right-hand limit is evaluated by factoring $\sqrt{x}$ from the numerator:
$\lim_{x \to 0^+} \frac{\sqrt{x^{2}+x} - \sqrt{x}}{x^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{x+1} - 1)}{x \sqrt{x}} = \lim_{x \to 0^+} \frac{\sqrt{x+1} - 1}{x}$
Multiplying by the conjugate $\frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}$ further yields:
$\lim_{x \to 0^+} \frac{(x+1) - 1}{x(\sqrt{x+1} + 1)} = \lim_{x \to 0^+} \frac{1}{\sqrt{x+1} + 1} = \frac{1}{2}$
Equating these results to $f(0) = q$, we find $p+2 = \frac{1}{2}$ and $q = \frac{1}{2}$, which results in $p = -\frac{3}{2}$ and $q = \frac{1}{2}$.

Rationalization is the most effective approach for resolving the indeterminate form encountered at $y=0$. Multiplying the numerator and denominator by the conjugate of the numerator, $\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}$, simplifies the numerator to $\sqrt{1+y^{4}}-1$. Next, rationalize this new numerator by multiplying the entire expression by $\frac{\sqrt{1+y^{4}}+1}{\sqrt{1+y^{4}}+1}$, which simplifies the numerator to $1+y^{4}-1$, or $y^{4}$. Since this $y^{4}$ cancels out the $y^{4}$ term in the denominator, you are left with the expression $\frac{1}{(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2})(\sqrt{1+y^{4}}+1)}$. Evaluating this at $y=0$ yields $\frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)}$, which simplifies to $\frac{1}{(\sqrt{2}+\sqrt{2})(2)}$, resulting in the final value of $\frac{1}{4\sqrt{2}}$.

Rationalization is the primary technique for resolving indeterminate forms involving radicals, as multiplying the numerator and denominator by the conjugate $\sqrt{x^{2}+2\sin x+1} + \sqrt{\sin^{2} x - x + 1}$ allows for the simplification of the denominator. Performing this operation transforms the expression into $\frac{(x+2\sin x)(\sqrt{x^{2}+2\sin x+1} + \sqrt{\sin^{2} x - x + 1})}{x^{2} + 2\sin x - \sin^{2} x + x}$. As $x$ approaches $0$, the dominant linear terms in the denominator reduce to $2\sin x + x$, while the numerator simplifies by substituting the small-angle approximation $\sin x \approx x$. Consequently, the denominator becomes $2x + x = 3x$, and the numerator becomes $(x + 2x)(\sqrt{1} + \sqrt{1})$, which equals $3x \cdot 2 = 6x$. Dividing these results, $\frac{6x}{3x}$, yields a final limit of $2$.

A function is continuous at any point if the left-hand limit, the right-hand limit, and the function value are all equivalent at that point. To determine if this function is continuous, we must ensure these conditions are met at the transition points $x=1$, $x=3$, and $x=5$. At $x=1$, the left-hand limit is $5$ and the right-hand limit is $a+b$, which necessitates $a+b=5$. At $x=5$, the left-hand limit is $b+5(5)$ and the function value is $30$, which leads to the equation $b+25=30$, meaning $b=5$.

Substituting $b=5$ into the first equation yields $a=0$. We must now verify if these values satisfy the continuity condition at $x=3$, where the left-hand limit is $a+3b$ and the right-hand limit is $b+5(3)$, or $b+15$. With $a=0$ and $b=5$, the left-hand limit results in $0+3(5)=15$, while the right-hand limit results in $5+15=20$. Because $15 \neq 20$, the function cannot be continuous at $x=3$ for any values of $a$ and $b$.

A function is continuous at a specific point if the limit as the variable approaches that point from the left and the right equals the function's actual value at that point. We determine the status of $f(x) = [x] - \left[ \frac{x}{4} \right]$ at $x=4$ by evaluating these three conditions to see if they converge to the same result.

When $x$ approaches 4 from the right, $[x]$ is 4 and $\left[ \frac{x}{4} \right]$ is 1, leading to a limit of $4 - 1 = 3$. Approaching from the left, $x$ is slightly less than 4, so $[x]$ evaluates to 3 and $\left[ \frac{x}{4} \right]$ becomes 0, which also results in $3 - 0 = 3$. Since the direct evaluation at $f(4)$ yields $4 - 1 = 3$, the limits from both sides match the function value, confirming that the function is continuous at $x=4$.

For the limit of this rational function to exist as a finite value, the numerator must vanish at $x=1$ since the denominator is already zero. Substituting $x=1$ into the numerator gives $1^{2}-a(1)+b=0$, which simplifies to $1-a+b=0$. To resolve the indeterminate form, differentiating the numerator and denominator with respect to $x$ yields the expression $\frac{2x-a}{1}$, which must equate to the limit value of 5 at $x=1$, resulting in the equation $2-a=5$.

Solving $2-a=5$ gives $a=-3$, and inserting this value into the previous equation $1-(-3)+b=0$ reveals that $b=-4$. Finally, adding these two coefficients together, $a+b$ equals $-3+(-4)$, which results in -7.

Rationalizing the numerator by multiplying by the conjugate $\sqrt{\pi}+\sqrt{2\sin^{-1}x}$ transforms the expression into $\frac{\pi - 2\sin^{-1}x}{\sqrt{1-x}(\sqrt{\pi}+\sqrt{2\sin^{-1}x})}$. Noting that $\frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$, the numerator simplifies to $2\cos^{-1}x$, while the term in the denominator $\sqrt{\pi}+\sqrt{2\sin^{-1}x}$ approaches $2\sqrt{\pi}$ as $x \to 1^{-}$, leaving the expression as $\frac{1}{2\sqrt{\pi}} \lim_{x\to 1^{-}} \frac{2\cos^{-1}x}{\sqrt{1-x}}$.

Substituting $x = \cos\theta$, where $\theta \to 0^{+}$ as $x \to 1^{-}$, transforms the remaining limit into $\frac{2\theta}{\sqrt{1-\cos\theta}} = \frac{2\theta}{\sqrt{2\sin^2(\theta/2)}} = \frac{2\theta}{\sqrt{2}\sin(\theta/2)}$. Applying the fundamental limit $\lim_{\theta\to 0} \frac{\theta}{\sin(\theta/2)} = 2$, the calculation yields $\frac{1}{2\sqrt{\pi}} \cdot 2\sqrt{2}$, which simplifies to $\sqrt{\frac{2}{\pi}}$.

A function $f(x)$ is continuous at $x=0$ if $\lim_{x \to 0} f(x) = f(0)$. Rewriting the function as $f(x) = \frac{e^{2x}-1 - x(k-1)}{x(e^{2x}-1)}$, the limit exists only if the numerator approaches zero as $x \to 0$. Using the Taylor series expansion for the exponential function, $e^{2x} = 1 + 2x + 2x^2 + \dots$, the expression becomes $\frac{1 + 2x + 2x^2 - 1 - kx + x}{x(2x + 2x^2)} = \frac{x(3-k) + 2x^2}{2x^2 + 2x^3}$.

To ensure a finite limit exists, the coefficient of the lowest power of $x$ in the numerator must be zero, as the denominator is of order $x^2$. Setting the numerator term $x(3-k)$ to zero yields $3-k=0$, implying $k=3$. Substituting $k=3$ into the simplified expression results in $\lim_{x \to 0} \frac{2x^2}{2x^2} = 1$, which defines the value $f(0)=1$. Thus, the ordered pair is $(k, f(0)) = (3, 1)$.

The substitution method simplifies limits containing radicals by converting the expression into a more manageable algebraic form. Defining $t = (27+x)^{1/3}$ implies that $t^3 = 27+x$, and as $x$ approaches $0$, $t$ approaches $3$. Using the standard form of this limit problem, which is $\lim_{x\to 0} \frac{(27+x)^{1/3}-3}{9-(27+x)^{2/3}}$, the expression is transformed into $\lim_{t\to 3} \frac{t-3}{9-t^2}$. Factoring the denominator as $9-t^2 = -(t^{2-}9) = -(t-3)(t+3)$, the expression becomes $\lim_{t\to 3} \frac{t-3}{-(t-3)(t+3)}$. Canceling the common factor $(t-3)$ results in $\lim_{t\to 3} -\frac{1}{t+3}$, which evaluates to $-\frac{1}{3+3} = -\frac{1}{6}$.