Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

Evaluating this limit involves isolating the rational component from the exponential terms and applying the standard limit definition $(1 + \frac{k}{u})^u \to e^k$ to simplify the behavior as $x$ approaches infinity. The rational expression $\frac{2x^{2-}3x+5}{3x^{2+}5x+4}$ simplifies to the constant $\frac{2}{3}$ because the leading terms $2x^2$ and $3x^2$ dominate the expression. The exponential portion is handled by factoring $3x$ out of the base terms $(3x-1)^{x/2}$ and $(3x+2)^{x/2}$, which cancels the $(3x)^{x/2}$ terms and yields:

$\frac{(1-\frac{1}{3x})^{x/2}}{(1+\frac{2}{3x})^{x/2}}$

Each factor in this expression takes the form $(1+\frac{a}{x})^{bx}$, where the limit is $e^{ab}$. Applying this rule, the numerator converges to $e^{(-1/3) \cdot (1/2)} = e^{-1/6}$, while the denominator converges to $e^{(2/3) \cdot (1/2)} = e^{1/3}$. Combining these parts leads to $\frac{2}{3} \times \frac{e^{-1/6}}{e^{1/3}}$, which simplifies further to $\frac{2}{3} \times e^{-1/6 - 1/3} = \frac{2}{3} \times e^{-1/2}$, resulting in $\frac{2}{3\sqrt{e}}$.

A function is continuous at a specific point if the left-hand limit, the right-hand limit, and the value of the function at that point are all equal. For the function to be continuous at $x=0$, the value at the point, $f(0)=4$, must be equal to both one-sided limits. Investigating the left-hand side where $x < 0$, we have $\lim_{x\to 0^-} \frac{2}{x} (\sin((k_1+1)x) + \sin((k_2-1)x))$. Applying the standard limit $\lim_{x\to 0} \frac{\sin(ax)}{x} = a$, this expression simplifies to $2(k_1+1 + k_2-1)$, which equals $2(k_1+k_2)$. Equating this to 4, we obtain the relation $k_1+k_2=2$.

For the right-hand side where $x > 0$, we consider $\lim_{x\to 0^+} \frac{2}{x} \ln \left( \frac{2+k_1 x}{2+k_2 x} \right)$, which can be expressed as $\lim_{x\to 0^+} 2 \left( \frac{\ln(1 + \frac{k_1}{2}x)}{x} - \frac{\ln(1 + \frac{k_2}{2}x)}{x} \right)$. Using the standard limit $\lim_{x\to 0} \frac{\ln(1+ax)}{x} = a$, this becomes $2(\frac{k_1}{2} - \frac{k_2}{2})$, reducing to $k_1-k_2=4$. Solving the system of equations $k_1+k_2=2$ and $k_1-k_2=4$ yields $k_1=3$ and $k_2=-1$, resulting in $k_1^{2+}k_2^2 = 3^2 + (-1)^2 = 10$.

The limit presents an indeterminate form of type $1^{\infty }$, which is evaluated using the identity $\lim_{x \to \infty } [f(x)]^{g(x)} = e^{\lim_{x \to \infty } g(x)(f(x)-1)}$. Identifying the base $f(x) = \left( \frac{e}{1-e} \right) \left( \frac{1}{e} - \frac{x}{1+x} \right)$ and the exponent $g(x) = x$, we subtract one from the base to initiate simplification. Distributing the term $\frac{e}{1-e}$ yields $\frac{1}{1-e} - \frac{ex}{(1-e)(1+x)} - 1$, and combining the constant terms $\frac{1}{1-e} - 1$ results in $\frac{e}{1-e}$. Factoring this out leaves $\frac{e}{1-e} \left( 1 - \frac{x}{1+x} \right)$, which simplifies further to $\frac{e}{1-e} \left( \frac{1}{1+x} \right)$.

Multiplying this result by the exponent $x$ and evaluating the limit as $x \to \infty$ gives $e^{\lim_{x \to \infty } \left( \frac{e}{1-e} \cdot \frac{x}{1+x} \right)} = e^{e/(1-e)}$, establishing $\alpha = e^{e/(1-e)}$ and consequently $\log_e \alpha = \frac{e}{1-e}$. Substituting this value into the final expression yields:
$\frac{\frac{e}{1-e}}{1 + \frac{e}{1-e}} = \frac{e}{(1-e) + e} = e$

The greatest integer function $\lfloor t \rfloor$ satisfies the property that as $x \to 0^{+}$, the product $x \lfloor \frac{k}{x} \rfloor$ approaches $k$ for any constant $k$. Applying this limit property to each term simplifies the entire expression into a sum of the first $p$ natural numbers minus the sum of the squares of the first nine natural numbers. The first part is $\frac{p(p+1)}{2}$ and the second part, calculated as $\frac{9(10)(19)}{6}$, is 285. The resulting inequality $\frac{p(p+1)}{2} - 285 \geq 1$ implies $p(p+1) \geq 572$. Since $23 \times 24 = 552$ and $24 \times 25 = 600$, the least natural number $p$ satisfying this condition is 24.

For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots is defined by the ratio $-B/A$. Applying this to the given equation, the sum $\alpha_t + \beta_t$ corresponds to $-\frac{(t+2)^{1/6}-1}{(t+2)^{1/7}-1}$. As $t$ approaches $-1$, the term $t+2$ approaches $1$. By substituting $u = t+2$, the limit of the sum simplifies to $\lim_{u\to 1} -\frac{u^{1/6}-1}{u^{1/7}-1}$.

This expression fits the standard limit form $\lim_{x\to 1} \frac{x^n-1}{x^m-1} = \frac{n}{m}$. Applying this property to the exponents $1/6$ and $1/7$, the limit $a+b$ evaluates to $-\frac{1/6}{1/7}$, which simplifies to $-7/6$. Squaring this result yields $49/36$. Multiplying this value by 72 gives the final result of 98.

To identify where the function $f(x) = \lfloor x \rfloor + |x-2|$ fails to be continuous or differentiable, we decompose the expression into piecewise segments within the interval $(-2, 3)$. Because the floor function $\lfloor x \rfloor$ remains constant between integers, the function behaves linearly within each sub-interval. Analyzing these segments reveals that the function exhibits jump discontinuities at each integer boundary $x = -1$, $x = 0$, $x = 1$, and $x = 2$, where the limit from the left side fails to match the function's value at those points.

Since a point of discontinuity is inherently a point of non-differentiability, the function lacks a derivative at all four of these integer locations. There are no additional points in the domain where the function is continuous but non-differentiable, establishing that the number of discontinuity points $m$ is $4$ and the number of non-differentiable points $n$ is $4$. Summing these counts leads to the result $m+n=8$.

Evaluating limits that result in an indeterminate form often requires using Maclaurin series expansions to compare coefficients of the variable $x$. Since the denominator behaves like $x^3$ as $x$ approaches $0$ because $\sin x$ is equivalent to $x$, the numerator must also be of the order $x^3$ for the limit to converge to a non-zero value, implying the coefficients of $x$ and $x^2$ in the numerator must be zero. Expanding the components gives $a x^2 (1 + x + \frac{x^2}{2} + \dots) - b(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots) + c x(1 - x + \frac{x^2}{2} - \dots)$, which simplifies to the expression
$(c-b)x + \left(a - c + \frac{b}{2}\right)x^2 + \left(a - \frac{b}{3} + \frac{c}{2}\right)x^3 + \cdots$
Setting the coefficients of $x$ and $x^2$ to zero yields $c - b = 0$ and $a - c + \frac{b}{2} = 0$, which implies $c = b$ and $a = \frac{b}{2}$. Substituting these values into the $x^3$ coefficient equation $a - \frac{b}{3} + \frac{c}{2} = 1$ leads to $\frac{b}{2} - \frac{b}{3} + \frac{b}{2} = 1$, or $\frac{2b}{3} = 1$, resulting in $b = \frac{3}{2}$, $c = \frac{3}{2}$, and $a = \frac{3}{4}$. Calculating the final value gives $16(a^2 + b^2 + c^2) = 16((\frac{3}{4})^2 + (\frac{3}{2})^2 + (\frac{3}{2})^2) = 16(\frac{9}{16} + \frac{9}{4} + \frac{9}{4}) = 9 + 36 + 36$, which equals $81$.

Continuity at a specific point requires the function's value to be perfectly aligned with the limits approaching that point from both directions. For the function to be continuous at $x = \frac{\pi}{2}$, the left-hand limit, the value $f(\frac{\pi}{2})$, and the right-hand limit must all be equal.

We start by evaluating the left-hand limit $\lim_{x \to \frac{\pi}{2}^-} (\frac{8}{7})^{\frac{\tan 8x}{\tan 7x}}$. As $x$ approaches $\frac{\pi}{2}$, the exponent $\frac{\tan 8x}{\tan 7x}$ tends toward 0, making the limit equivalent to $(\frac{8}{7})^0 = 1$. Since the function is continuous, this must equal the value defined at the point, $a - 8$, which gives $a - 8 = 1$, and consequently $a = 9$.

Next, we address the right-hand limit $\lim_{x \to \frac{\pi}{2}^+} (1 + |\cot x|)^{\frac{b}{a}|\tan x|}$. This expression takes the form of the standard limit $\lim_{u \to 0} (1+u)^{k/u} = e^k$, where $u = |\cot x|$ and $k = \frac{b}{a}$. The limit evaluates to $e^{\frac{b}{a}}$, and setting this equal to the previously determined value of 1 gives $e^{\frac{b}{a}} = 1$. This implies $\frac{b}{a} = 0$, leading to $b = 0$. Finally, the calculation of $a^2 + b^2$ becomes $9^2 + 0^2$, which results in 81.

Because the derivative $f'(x) = \frac{1}{2} - \frac{2}{x^2}$ remains negative throughout the interval $(0, 2)$, the function $f(x)$ is strictly decreasing, which ensures that the minimum value of $f(t)$ on the interval $(0, x]$ for any $x \le 1$ is simply the value at the right endpoint, $f(x) = \frac{x}{2} + \frac{2}{x}$. This establishes the first part of $g(x)$, while the second part is defined as $\frac{3}{2} + x$ for $1 < x < 2$.

Evaluating the junction at $x=1$ reveals continuity because the function value $g(1) = \frac{1}{2} + 2 = \frac{5}{2}$ matches both the left-hand limit $\lim_{x \to 1^-} (\frac{x}{2} + \frac{2}{x}) = \frac{5}{2}$ and the right-hand limit $\lim_{x \to 1^+} (\frac{3}{2} + x) = \frac{3}{2} + 1 = \frac{5}{2}$.

Differentiability at $x=1$ depends on the equality of one-sided derivatives. The left-hand derivative is $f'(1) = \frac{1}{2} - 2 = -\frac{3}{2}$, while the derivative of the right-hand branch is the constant $1$. Since these values are not equal, the function is not differentiable at this point.

Simplifying these indeterminate limits involves rationalizing radical expressions and applying trigonometric identities. For $a$, multiplying the numerator and denominator by the conjugate $\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}$ removes the outer square root, leading to $\frac{\sqrt{1+x^4}-1}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})}$. Multiplying again by the conjugate $\sqrt{1+x^4}+1$ clears the inner radical, which results in $\frac{x^4}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)}$. Canceling the $x^4$ terms and evaluating the limit as $x \to 0$ gives $a = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.

For $b$, substituting $\sin^2 x$ with $1-\cos^2 x$ provides the product $(1-\cos x)(1+\cos x)$ in the numerator. Rationalizing the denominator with the conjugate $\sqrt{2}+\sqrt{1+\cos x}$ yields the expression $\frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$. Canceling the $1-\cos x$ term simplifies the limit to $(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$. Applying the limit as $x \to 0$ gives $b = (2)(2\sqrt{2}) = 4\sqrt{2}$. Calculating the final value $ab^3$ leads to $\frac{1}{4\sqrt{2}} \times (4\sqrt{2})^3$, which simplifies to $(4\sqrt{2})^2 = 32$.

For the given limit to equal $1/3$, the numerator must approach zero as $x$ approaches 0, because the denominator $3\tan^2 x$ also approaches 0. Using Taylor series expansions about $x=0$, the expressions can be approximated as $\sin x \approx x - \frac{x^3}{6}$, $\cos x \approx 1 - \frac{x^2}{2}$, and $\log_e(1-x) \approx -x - \frac{x^2}{2} - \frac{x^3}{3}$. Substituting these into the numerator expression yields
$3 + \alpha\left(x - \frac{x^3}{6}\right) + \beta\left(1 - \frac{x^2}{2}\right) + \left(-x - \frac{x^2}{2} - \frac{x^3}{3}\right)$
Regrouping this expression by the powers of $x$ gives $(3+\beta) + (\alpha-1)x + (-\frac{\beta}{2} - \frac{1}{2})x^2 + \dots$. Since the denominator $3\tan^2 x$ behaves like $3x^2$ near 0, the coefficients of the constant and linear terms in the numerator must be zero to ensure the limit is a non-zero finite value.

Solving $3+\beta = 0$ results in $\beta = -3$, and solving $\alpha-1 = 0$ results in $\alpha = 1$. With these values, the coefficient of the $x^2$ term becomes $(-\frac{-3}{2} - \frac{1}{2})$, which simplifies to $1$. The limit of the ratio is then $\frac{1x^2}{3x^2}$, which equals $1/3$, confirming these coefficients are correct. Finally, evaluating the requested expression $2\alpha - \beta$ gives $2(1) - (-3)$, which results in 5.

A function is continuous at a specific point if the left-hand limit, the right-hand limit, and the value of the function at that point are all identical. For this function to be continuous at $x=0$, we first determine the limit from the left by applying the trigonometric identity $1-\cos 2x = 2\sin^2 x$. This allows us to write
$\lim_{x \to 0^{-}} \frac{2\sin^2 x}{x^2} = 2$
Since the value of the function at $x=0$ is defined as $\alpha$, we must have $\alpha = 2$.

For the limit from the right, we use the identity $1-\cos x = 2\sin^2(x/2)$, which transforms the expression into $\beta \frac{\sqrt{2}\sin(x/2)}{x}$. Simplifying this limit, we observe
$\lim_{x \to 0^{+}} \frac{\beta \sqrt{2} \sin(x/2)}{2(x/2)} = \frac{\beta}{\sqrt{2}}$
Equating this to the previously established limit of $2$ yields $\frac{\beta}{\sqrt{2}} = 2$, which simplifies to $\beta = 2\sqrt{2}$. Finally, calculating the sum of the squares gives $\alpha^2 + \beta^2 = 2^2 + (2\sqrt{2})^2 = 4 + 8 = 12$.

The greatest integer function $\lfloor u \rfloor$ is discontinuous precisely where the argument $u$ takes on an integer value, meaning the function $f(x) = 2x^2 + x + \lfloor x^2 \rfloor - \lfloor x \rfloor$ can only fail to be continuous at points where $x$ or $x^2$ is an integer. Within the interval $[-1, 2]$, these conditions yield a set of potential discontinuity points at $\{-1, 0, 1, \sqrt{2}, \sqrt{3}, 2\}$.

Testing the function at $x = -1$ reveals a jump discontinuity because the right-hand limit is $2$ while the function value is $3$, and evaluating the function at $x = 0$ confirms a similar discontinuity with a left-hand limit of $1$ and a function value of $0$. Investigating $x = 1$ shows that the left-hand limit, right-hand limit, and function value are all $3$, making the function continuous there, while a similar evaluation at $x = 2$ confirms continuity with a value of $12$. Finally, the values $x = \sqrt{2}$ and $x = \sqrt{3}$ also serve as points of discontinuity, as the $\lfloor x^2 \rfloor$ term introduces a jump as $x^2$ crosses an integer at these specific locations. Counting these occurrences leads to a total of 4 points where the function is not continuous.

To evaluate the limit of the function $f(x) = \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$ at $x=0$, we must account for the behavior of the fractional part function $\{x\}$ as it approaches zero from different sides. For the right-hand limit, where $x \to 0^+$, the fractional part $\{x\}$ is equivalent to $x$. Substituting this, we obtain $\lim_{h \to 0^+} \frac{\cos^{-1}(1-h^2) \sin^{-1}(1-h)}{h-h^3}$. As $h$ approaches zero, $\sin^{-1}(1-h)$ approaches $\sin^{-1}(1) = \frac{\pi}{2}$, and the factor $1-h^2$ in the denominator approaches $1$. The expression simplifies to $\frac{\pi}{2} \lim_{h \to 0} \frac{\cos^{-1}(1-h^2)}{h}$. By setting $\theta = \cos^{-1}(1-h^2)$, we get $\cos \theta = 1-h^2$, which implies $2\sin^2(\frac{\theta}{2}) = h^2$, or $\theta \approx \sqrt{2}h$. Thus, the right-hand limit $R$ is $\frac{\pi}{2} \cdot \sqrt{2} = \frac{\pi}{\sqrt{2}}$.

For the left-hand limit, where $x \to 0^-$, the fractional part $\{x\}$ behaves as $1+x$, which can be written as $1-h$ for a small positive value $h$. Substituting this into the function yields $\lim_{h \to 0^+} \frac{\cos^{-1}(1-(1-h)^2) \sin^{-1}(1-(1-h))}{(1-h)-(1-h)^3}$. Expanding the terms, the numerator becomes $\cos^{-1}(2h-h^2) \sin^{-1}(h)$, which approaches $\cos^{-1}(0) \cdot h = \frac{\pi}{2}h$ as $h$ goes to zero. The denominator $(1-h)(1-(1-h)^2)$ simplifies to $(1-h)(2h-h^2)$, which is approximately $2h$ when $h$ is very small. Consequently, the left-hand limit $L$ becomes $\lim_{h \to 0} \frac{(\frac{\pi}{2}h)}{2h} = \frac{\pi}{4}$.

With these values, we determine the final result by evaluating the expression $\frac{32}{\pi^2} (L^{2+}R^2)$. Substituting $L^2 = \frac{\pi^2}{16}$ and $R^2 = \frac{\pi^2}{2}$, the sum is $\frac{\pi^2}{16} + \frac{\pi^2}{2} = \frac{\pi^2 + 8\pi^2}{16} = \frac{9\pi^2}{16}$. Multiplying this sum by $\frac{32}{\pi^2}$ gives $\frac{32}{\pi^2} \cdot \frac{9\pi^2}{16} = 2 \cdot 9$, which equals 18.

The behavior of this expression near zero is dominated by the small-angle approximation $\cos \theta \approx 1 - \frac{\theta^2}{2}$ combined with the binomial expansion $(1 - u)^k \approx 1 - ku$. By treating each term in the product as $(\cos nx)^{1/n}$, each factor becomes approximately $(1 - \frac{n^2x^2}{2})^{1/n}$, which simplifies to $1 - \frac{1}{n} \cdot \frac{n^2x^2}{2} = 1 - \frac{nx^2}{2}$.

The entire product can then be approximated as $(1 - \frac{1x^2}{2})(1 - \frac{2x^2}{2}) \dots (1 - \frac{10x^2}{2})$. Retaining only the first-order terms in $x^2$, this expands to $1 - (\frac{1}{2} + \frac{2}{2} + \dots + \frac{10}{2})x^2$. Replacing this back into the original formula yields $\lim_{x\to 0} 2 \left( \frac{1 - (1 - (\sum_{n=1}^{10} \frac{n}{2})x^2)}{x^2} \right)$, which simplifies to $2 \left( \frac{1}{2} + \frac{2}{2} + \dots + \frac{10}{2} \right)$. Evaluating the sum of integers from 1 to 10 as $\frac{10 \times 11}{2}$ confirms the final result is 55.

The behavior of the expression as $n$ approaches infinity is determined by the highest-degree terms in both the numerator and the denominator. The numerator involves the summation $\sum_{r=1}^{n-1} (-r^3 + r^2(n+1) - nr)$. Using standard power sum formulas, the sum $\sum_{r=1}^{n-1} r^3$ is $\frac{(n-1)^2 n^2}{4}$, which approximates to $\frac{n^4}{4}$, and the sum $\sum_{r=1}^{n-1} r^2(n+1)$ approximates to $\frac{n^4}{3}$. Combining these, the leading term of the numerator is $\frac{n^4}{12}$. For the denominator, the expression $\sum_{r=1}^{n} r^3 - \sum_{r=1}^{n} r^2$ is dominated by the cubic sum $\sum_{r=1}^n r^3$, which approximates to $\frac{n^4}{4}$ because the quadratic sum $\sum_{r=1}^n r^2$ contributes terms of lower order. Dividing the numerator's leading coefficient $\frac{1}{12}$ by the denominator's leading coefficient $\frac{1}{4}$ yields the final limit of $\frac{1}{3}$.

The continuity of $f(x) = \left[ \frac{x}{2}+3 \right] - \left[ \sqrt{x} \right]$ relies on the points where these component greatest integer functions jump. For $\left[ \frac{x}{2}+3 \right]$, jumps occur when the argument is an integer, leading to discontinuities at $x=2, 4, 6,$ and $8$, while $\left[ \sqrt{x} \right]$ jumps when $\sqrt{x}$ is an integer, occurring at $x=1$ and $x=4$. Because both functions experience a unit jump at $x=4$, their difference remains constant and the function is continuous at that point. Therefore, the function is discontinuous at the remaining points $x \in \{1, 2, 6, 8\}$, and summing these yields $1+2+6+8=17$.

The limit expression $\lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t-x} = 1$ is a derivative-based structure that characterizes the function $f(x)$. Applying L'Hôpital's rule by differentiating the numerator and denominator with respect to $t$ gives $\frac{2t f(x) - x^2 f'(t)}{1}$. Setting $t$ equal to $x$ simplifies this to the linear first-order differential equation $2x f(x) - x^2 f'(x) = 1$.

Rearranging the equation into the standard form $f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}$ allows for the application of an integrating factor, calculated as $e^{\int -\frac{2}{x} dx} = \frac{1}{x^2}$. Multiplying the entire equation by this factor results in $\frac{d}{dx} \left( \frac{f(x)}{x^2} \right) = -\frac{1}{x^4}$. Integrating both sides with respect to $x$ yields $\frac{f(x)}{x^2} = \frac{1}{3x^3} + C$.

Given the condition $f(1) = 1$, substituting $x=1$ into the equation results in $1 = \frac{1}{3} + C$, which identifies the constant as $C = \frac{2}{3}$. This defines the function as $f(x) = \frac{2x^{3+}1}{3x}$. Substituting the necessary values produces $f(2) = \frac{17}{6}$ and $f(3) = \frac{55}{9}$. Finally, the expression $2 f(2) + 3 f(3)$ evaluates to $2 \left( \frac{17}{6} \right) + 3 \left( \frac{55}{9} \right) = \frac{17}{3} + \frac{55}{3}$, resulting in 24.