Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

The expression $\lim_{x\to 0} (\frac{\tan x}{x})^{\frac{1}{x^2}}$ presents an indeterminate $1^{\infty }$ form, which is effectively resolved by utilizing the limit identity $\lim_{x\to a} [f(x)]^{g(x)} = e^{\lim_{x\to a} [f(x)-1]g(x)}$. Applying this approach to the given function leads to $p = e^{\lim_{x\to 0} (\frac{\tan x}{x} - 1) \frac{1}{x^2}}$, which simplifies to $e^{\lim_{x\to 0} \frac{\tan x - x}{x^3}}$.

Substituting the standard Taylor series expansion for $\tan x$, expressed as $x + \frac{x^3}{3} + \frac{2}{15}x^5 + \dots$, into the numerator causes the leading $x$ terms to cancel, resulting in $e^{\lim_{x\to 0} \frac{\frac{x^3}{3} + \dots}{x^3}}$. This limit evaluates to $e^{1/3}$, which means that $\log_e p = \frac{1}{3}$. Multiplying this result by $96$ yields $32$.

The core concept behind this problem is the telescoping sum method, where a series is expressed in a form that causes consecutive intermediate terms to cancel out, leaving only the first and last components. By identifying the general term of the series, one can apply partial fraction decomposition to transform the series into a difference of two terms.

The provided sum of the first $n$ terms is $S_n = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$. The general term $T_n$ is obtained by subtracting the sum of the first $n-1$ terms from the sum of the first $n$ terms, given by $T_n = S_n - S_{n-1}$. Calculating this difference, we find $T_n = \frac{(2n-1)(2n+1)(2n+3)(2n+5) - (2n-3)(2n-1)(2n+1)(2n+3)}{64}$, which simplifies to $T_n = \frac{(2n-1)(2n+1)(2n+3)}{8}$ after factoring out the common terms and simplifying the expression in the numerator.

The reciprocal of the general term is $\frac{1}{T_n} = \frac{8}{(2n-1)(2n+1)(2n+3)}$. This expression can be decomposed using partial fractions into $2 \left( \frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)} \right)$. When summing this from $r=1$ to $n$, the series becomes $2 \left( \left( \frac{1}{1 \times 3} - \frac{1}{3 \times 5} \right) + \left( \frac{1}{3 \times 5} - \frac{1}{5 \times 7} \right) + \dots + \left( \frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)} \right) \right)$.

All intermediate terms cancel out during the summation, leaving only the initial part of the first term and the final part of the last term: $2 \left( \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right)$. As $n$ approaches infinity, the fraction $\frac{1}{(2n+1)(2n+3)}$ vanishes, and the value of the series converges to $2 \times \frac{1}{3} = \frac{2}{3}$.

The function $f(x) = \max\{x, x^3, x^5, \dots, x^{21}\}$ is defined by different power functions across distinct intervals of $x$. Analyzing these intervals reveals that for $x > 1$, $x^{21} > x^{19} > \dots > x$, so $f(x) = x^{21}$. For $0 \le x \le 1$, the function follows $f(x) = x$ because $x \ge x^3 \ge \dots \ge x^{21}$. For $-1 < x < 0$, the function is $f(x) = x^{21}$ because the values are all negative and $x^{21}$ has the smallest absolute value. For $x \le -1$, $f(x) = x$ as $|x| \ge |x^k|$ for all $k \in \{3, 5, \dots, 21\}$. By checking the limits at the intersection points $x=1, x=0$, and $x=-1$, we find the left-hand and right-hand limits coincide at each point, confirming the function is continuous everywhere, so $n=0$.

Differentiability at the transition points is determined by the equality of the one-sided derivatives. At $x=1$, the derivative of the left side is $1$ and the right side is $\frac{d}{dx}(x^{21})|_{x=1} = 21(1)^{20} = 21$, which are unequal. At $x=0$, the derivative of the left side is $\frac{d}{dx}(x^{21})|_{x=0} = 21(0)^{20} = 0$ and the right side is $1$, which are also unequal. At $x=-1$, the derivative of the left side is $1$ and the right side is $\frac{d}{dx}(x^{21})|_{x=-1} = 21(-1)^{20} = 21$, which are again unequal. These three points constitute the set of points where the function is not differentiable, so $m=3$. Therefore, $m+n = 3+0 = 3$.

The provided limit expression follows the form $1^{\infty }$ since $f(2)=1$, which allows us to use the standard property $\lim_{x\to 0} [g(x)]^{h(x)} = e^{\lim_{x\to 0} h(x)[g(x)-1]}$. By substituting $g(x)=f(2+x)$ and $h(x)=\frac{3}{x}$, we obtain $e^{\lim_{x\to 0} \frac{3}{x}(f(2+x)-1)}$. Because $f(2)=1$, we can rewrite the term $(f(2+x)-1)$ as $(f(2+x)-f(2))$, transforming the limit into $e^{3 \lim_{x\to 0} \frac{f(2+x)-f(2)}{x}}$. This derivative definition yields $e^{3f'(2)}$, and given $f'(2)=4$, the exponent evaluates to 12, implying $\alpha=12$.

Substituting $\alpha=12$ into the polynomial expression results in $y=4x^{3-}4x^{2-}4(12-7)x-12$, which simplifies to $y=4x^{3-}4x^{2-}20x-12$. Factoring this cubic equation, we extract the common factor of 4 to get $4(x^{3-}x^{2-}5x-3)$. Through algebraic manipulation, we find that $(x+1)$ is a repeated factor, leading to the fully factored form $y=4(x+1)^2(x-3)$. The roots of the polynomial are therefore $-1$ and $3$, indicating that the curve intersects the x-axis at exactly two points.

The differentiability of $f(x)$ for all $x$ requires it to be continuous and possess a continuous derivative at the transition point $x=1$. Enforcing continuity at $x=1$ requires the values $-3a-2$ and $a^{2+}b$ to be equal, while the requirement for equal derivatives on both sides implies that the derivative of $-3ax^{2-}2$, which is $-6ax$, must equal $b$ at $x=1$. Substituting $b=-6a$ into the continuity condition $-3a-2=a^{2+}b$ yields $a^{2-}3a+2=0$, which factors as $(a-1)(a-2)=0$. Since the problem specifies $a>1$, we must have $a=2$, which then determines $b=-12$.

With the specific parameters determined, the function becomes $f(x)=-6x^{2-}2$ for $x<1$ and $4-12x$ for $x \ge 1$. To find the enclosed region bounded by $y=-20$, we identify the limits of integration by setting $f(x)=-20$, solving $-6x^{2-}2=-20$ to find the lower bound $x=-\sqrt{3}$ and $4-12x=-20$ to find the upper bound $x=2$. The total area is defined by the sum of integrals $\int_{-\sqrt{3}}^{1} (-6x^{2-}2 - (-20)) \, dx + \int_{1}^{2} (4-12x - (-20)) \, dx$. Evaluating the first integral produces $[-2x^{3+}18x]_{-\sqrt{3}}^{1} = 16+12\sqrt{3}$, and the second yields $[24x-6x^2]_1^2 = 6$, resulting in a total area of $22+12\sqrt{3}$. Consequently, $\alpha=22$ and $\beta=12$, which leads to $\alpha+\beta=34$.

Evaluating these limits relies on Taylor series expansions for polynomial approximations and standard properties for resolving indeterminate exponential forms. Regarding the first limit, representing $\tan^{-1} x$ as $x - \frac{x^3}{3} + \frac{x^5}{5}$ and the logarithmic component as $\frac{1}{2}(\ln(1+x) - \ln(1-x))$ allows for a series expansion centered at zero. When these functions are expanded to the fifth power, all terms of lower degree cancel out with the $-2x$ present in the numerator, leaving only the fifth-order terms to sum to $\frac{2}{5}x^5$, which confirms that the limit is indeed $\frac{2}{5}$.

The second limit presents an indeterminate form of the type $1^\infty$, which is best managed by substituting $t = x-1$ to utilize the fundamental property $\lim_{t \to 0} (1+t)^{1/t} = e$. This transformation rewrites the exponent as $\frac{2}{-t}$, effectively converting the expression into $((1+t)^{\frac{1}{t}})^{-2}$. As $t$ approaches zero, the inner expression approaches $e$, leading to the final result of $e^{-2}$, or $\frac{1}{e^2}$, validating that the second statement is also true.

Treating the provided relation as part of a system of equations requires substituting $x$ with $1/x$ to isolate the function $f(x)$. The original expression $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$ leads to a second equation, $f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2}$, when the substitution is applied. By multiplying the second equation by $6$ to obtain $6f\left(\frac{1}{x}\right) - 36f(x) = 70x - 15$ and adding this to the original expression, the $f\left(\frac{1}{x}\right)$ terms cancel out to yield $-35f(x) = \frac{35}{3x} + 70x - \frac{35}{2}$. Simplifying this results in $f(x) = -\frac{1}{3x} - 2x + \frac{1}{2}$. Inserting this into the limit expression $\lim_{x \to 0} \left( \frac{1}{\alpha x} + f(x) \right) = \beta$ produces $\lim_{x \to 0} \left( \frac{1}{\alpha x} - \frac{1}{3x} - 2x + \frac{1}{2} \right) = \beta$. For this limit to be a finite value $\beta$, the terms containing $1/x$ must eliminate each other, which occurs when $\alpha = 3$. With those terms removed, the limit evaluates to $\beta = 1/2$, resulting in $\alpha + 2\beta = 3 + 2\left(\frac{1}{2}\right) = 4$.

Evaluating this limit is straightforward when utilizing the standard approximations $\lim_{u\to 0} \frac{\tan u}{u} = 1$, $\lim_{u\to 0} \frac{\ln(1+u)}{u} = 1$, $\lim_{u\to 0} \frac{\tan^{-1} u}{u} = 1$, and $\lim_{u\to 0} \frac{e^u-1}{u} = 1$. As $x$ approaches $0^{+}$, each component in the expression can be replaced by its leading term: $\tan(5x^{1/3})$ approximates to $5x^{1/3}$, $\ln(1+3x^2)$ approximates to $3x^2$, $(\tan^{-1}(3\sqrt{x}))^2$ becomes $(3\sqrt{x})^2$ which is $9x$, and $(e^{5x^{4/3}}-1)$ behaves like $5x^{4/3}$.

Substituting these expressions into the numerator and denominator results in $\frac{5x^{1/3} \cdot 3x^2}{9x \cdot 5x^{4/3}}$. Multiplying the terms in the numerator gives $15x^{1/3 + 2}$, which is $15x^{7/3}$, while the denominator simplifies to $45x^{1 + 4/3}$, which is $45x^{7/3}$. Since the powers of $x$ are identical, they cancel out, leaving the ratio 15/45, which simplifies to 1/3.

Continuity at $x=0$ requires that the left-hand limit, the right-hand limit, and the function value $f(0)$ all converge to the same value. Starting with the right-hand limit, $\lim_{x \to 0^+} \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2}$, we note that for the limit to be a finite real number, the denominator must approach zero as $x$ approaches $0$. This condition dictates that $(0+c)^{1/3} - 2 = 0$, leading to $c=8$. Using L'Hopital's rule on the $\frac{0}{0}$ form, we differentiate the numerator and denominator to get $\lim_{x \to 0} \frac{\frac{1}{2}(x+4)^{-1/2}}{\frac{1}{3}(x+c)^{-2/3}}$. Substituting $x=0$ and $c=8$ evaluates this expression to $\frac{1/4}{1/12}$, which simplifies to $3$. Since this limit must equal $f(0) = 1+b$, we find $1+b=3$, so $b=2$.

Turning to the left-hand limit, the expression $\lim_{x \to 0^-} (1+ax)^{1/x}$ is a standard form that evaluates to $e^a$. Equating this to the previously established value of $3$ gives $e^a = 3$. Therefore, the product $e^a b c$ is calculated as $3 \times 2 \times 8$, which equals $48$.

The functional equation $f(2x) - f(x) = x$ can be rewritten as $f(x) - f\left(\frac{x}{2}\right) = \frac{x}{2}$. By replacing $x$ with $\frac{x}{2^{k-1}}$, the equation becomes $f\left(\frac{x}{2^{k-1}}\right) - f\left(\frac{x}{2^k}\right) = \frac{x}{2^k}$. Summing this expression from $k=1$ to $n$ leads to the telescoping series $f(x) - f\left(\frac{x}{2^n}\right) = x \sum_{k=1}^n \left(\frac{1}{2}\right)^k$, which simplifies to $f(x) - f\left(\frac{x}{2^n}\right) = x\left(1 - \frac{1}{2^n}\right)$.

Taking the limit as $n \to \infty$ gives G(x) = \lim_{n \to \infty } \left[ x\left(1 - \frac{1}{2^n}\right) \right]\ = x$. The required \sum is$\sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2$. Using the formula$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$, the \sum for$n=10$is$\frac{10(11)(21)}{6} = 385$.

To evaluate the limit, substitute $t = x - 1$, which transforms the expression into the form $\lim_{t \to 0} \frac{t(6 + \lambda \cos t) - \mu \sin t}{t^3} = -1$. Replacing $\cos t$ and $\sin t$ with their respective Taylor series expansions $1 - \frac{t^2}{2}$ and $t - \frac{t^3}{6}$ yields the numerator $6t + \lambda t(1 - \frac{t^2}{2}) - \mu(t - \frac{t^3}{6}) = (6 + \lambda - \mu)t + (\frac{\mu}{6} - \frac{\lambda}{2})t^3$. For the limit to exist and equal $-1$, the coefficient of $t$ must vanish, giving the equation $6 + \lambda - \mu = 0$, or $\mu - \lambda = 6$. Equating the coefficient of $t^3$ to $-1$ results in $\frac{\mu}{6} - \frac{\lambda}{2} = -1$, which simplifies to $\mu - 3\lambda = -6$. Solving the system formed by $\mu - \lambda = 6$ and $\mu - 3\lambda = -6$ yields $\lambda = 6$ and $\mu = 12$. Consequently, $\lambda + \mu = 6 + 12 = 18$.

The summation is a telescoping series, where terms are structured to cancel out upon expansion. Each term within the sum, $\frac{\tan\left(\frac{x}{2^{r+1}}\right) + \tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}$, is equivalent to the identity $\tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right)$. Summing these from $r=0$ to $n$ leads to $\tan(x) - \tan\left(\frac{x}{2^{n+1}}\right)$. As $n$ approaches infinity, $\tan\left(\frac{x}{2^{n+1}}\right)$ approaches 0, which simplifies the function to $f(x) = \tan(x)$.

The final limit, $\lim_{x\to 0} \frac{e^x - e^{\tan x}}{x - \tan x}$, is evaluated by rewriting the numerator as $e^{\tan x}(e^{x-\tan x} - 1)$. This transformation allows the expression to be calculated as the product $\lim_{x\to 0} e^{\tan x} \cdot \lim_{x\to 0} \frac{e^{x-\tan x} - 1}{x - \tan x}$. Since the limit of $e^{\tan x}$ is 1 and the term $\frac{e^{x-\tan x} - 1}{x - \tan x}$ reaches 1 according to the standard form $\lim_{u\to 0} \frac{e^u - 1}{u} = 1$, the final answer is 1.

A function is continuous at a point if its value at that point is equal to the limit of the function as it approaches that point. Evaluating the limit
$\lim_{x \to 0} \frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$
requires analyzing the small-angle approximations, where $\tan x \approx x + \frac{x^3}{3}$ and $\sin x \approx x - \frac{x^3}{6}$. These expansions reveal that the denominator $\tan x - \sin x$ behaves as $\frac{x^3}{2}$, while the numerator $\tan(\tan x) - \sin(\sin x)$ approximates to $x^3$. Substituting these leading terms into the expression yields $\frac{x^3}{x^3/2}$, which simplifies to $2$, the required value for $f(0)$.

For this limit to be finite, the numerator must approach zero at a rate that allows the $x^4$ term in the denominator to divide out all lower-order terms, which requires the constant term and the $x^2$ term in the Taylor expansion of the numerator to vanish. Expanding the trigonometric functions gives the numerator as $(1 - \frac{(2x)^2}{2!} + \dots) + a(1 - \frac{(4x)^2}{2!} + \dots) - b$, which simplifies to $(1 + a - b) - (2 + 8a)x^2 + \dots$ when organized by powers of $x$. Setting the constant term $1 + a - b$ equal to zero and the $x^2$ coefficient $-2 - 8a$ equal to zero provides the necessary conditions for the limit to exist. Solving the second equation yields $a = -1/4$, which leads to $b = 3/4$ when substituted into the first equation. Adding these two results together produces $a + b = 1/2$.

When evaluating a limit that approaches a finite value while the denominator simultaneously tends to zero, the numerator is required to also tend to zero to satisfy the necessary indeterminate form. As $x$ approaches $0$, the denominator expression $\sin 2x - \beta x$ evaluates to $0$, which forces the term $(\gamma - 1) e^{x^2}$ in the numerator to vanish, uniquely determining that $\gamma = 1$. Having removed this constant term, the expression simplifies to $\lim_{x \to 0} \frac{x^2 \sin(\alpha x)}{\sin 2x - \beta x} = 3$.

To solve for the remaining constants, we apply the standard Taylor series approximation $\sin \theta \approx \theta - \frac{\theta^3}{6}$ to the trigonometric terms in the numerator and denominator. This expansion results in $\lim_{x \to 0} \frac{x^2(\alpha x)}{(2x - \frac{8x^3}{6}) - \beta x} = 3$, which effectively rearranges to $\lim_{x \to 0} \frac{\alpha x^3}{(2 - \beta)x - \frac{4}{3}x^3} = 3$. For this expression to converge to a non-zero finite number, the linear coefficient in the denominator must be zero, requiring $2 - \beta = 0$, or $\beta = 2$. With this condition established, the limit becomes $\frac{\alpha x^3}{-\frac{4}{3} x^3} = 3$, which yields $\alpha = -4$. Calculating the requested sum, $\beta + \gamma - \alpha$ results in $2 + 1 - (-4) = 7$.

The expression $\lim_{x \to 0} \csc x \left( \sqrt{2 \cos^2 x + 3 \cos x} - \sqrt{\cos^2 x + \sin x + 4} \right)$ presents as an indeterminate $0 \cdot \infty$ form because $\csc x$ is the reciprocal of $\sin x$ and the radical expression simplifies to $\sqrt{2(1)^2 + 3(1)} - \sqrt{1^2 + 0 + 4} = \sqrt{5} - \sqrt{5} = 0$. To resolve this, rewrite the expression as a fraction with $\sin x$ in the denominator and multiply both the numerator and denominator by the conjugate $\sqrt{2 \cos^2 x + 3 \cos x} + \sqrt{\cos^2 x + \sin x + 4}$. This algebraic step transforms the numerator into $\cos^2 x + 3 \cos x - \sin x - 4$ and places the sum of the square roots in the denominator alongside $\sin x$.

As $x$ approaches $0$, the sum of the square roots in the denominator approaches $2\sqrt{5}$, allowing this term to be factored out of the limit as a constant $\frac{1}{2\sqrt{5}}$. We are then left to evaluate the limit of $\frac{\cos^2 x + 3 \cos x - \sin x - 4}{\sin x}$ as $x \to 0$. Applying L'Hopital's rule, we take the derivative of the numerator to get $-\sin 2x - 3 \sin x - \cos x$ and the derivative of the denominator to get $\cos x$. Evaluating these derivatives at $x=0$ yields $\frac{0 - 0 - 1}{1}$, which is $-1$. Multiplying this result by the previously factored constant $\frac{1}{2\sqrt{5}}$ produces a final value of $-\frac{1}{2\sqrt{5}}$.

Continuity and differentiability are determined by analyzing the behavior of the piecewise function at the boundaries where its definition changes. By simplifying the expression using the properties of the greatest integer function, the function is defined as $f(x)=3x$ for $x < 0$, $f(x)=x$ for $0 \le x < 1$, $f(x)=x+2$ for $1 \le x < 2$, and $f(x)=5$ for $x > 2$.

Testing continuity at the junction points $x=0$, $x=1$, and $x=2$ reveals the nature of the function's transition. At $x=0$, the limit from the left matches the value of the function from the right, ensuring continuity. However, at $x=1$, the left limit is $1$ while the right limit is $3$, and at $x=2$, the left limit is $4$ while the right limit is $5$. These two jump discontinuities establish that $\alpha=2$. Since differentiability requires continuity, the function is automatically not differentiable at $x=1$ and $x=2$. Additionally, examining $x=0$ shows the left-hand derivative is $3$ and the right-hand derivative is $1$, which are unequal, confirming the function is also not differentiable at $x=0$. Consequently, there are three points where the function is not differentiable, giving $\beta=3$. Summing these values yields $\alpha+\beta=5$.

A function involving the greatest integer operator $[u(x)]$ is discontinuous specifically at points where the expression inside the brackets yields an integer value. For the function f(x) = \[\frac{x^2}{2}]$ - [\sqrt{x}]$on the interval$x \in [0, 4]$, identifying the points of discontinuity requires finding where either$\frac{x^2}{2}$or$\sqrt{x}$ equals an integer.

Evaluating the first term, $\frac{x^2}{2}$ results in integers at the points $x = \sqrt{2}, x = 2, x = \sqrt{6}, x = 2\sqrt{2}, x = \sqrt{10}, x = 2\sqrt{3}, x = \sqrt{14}, x = 4$. For the second term, $\sqrt{x}$ hits integer values when $x = 1$ and $x = 4$. Aggregating these provides the complete set of candidate points where $x$ could potentially be discontinuous.

Testing each of these candidates reveals that at $x = 4$, the left-hand limit equals the function value, rendering the function continuous at the boundary. Eliminating this endpoint leaves exactly 8 distinct values, specifically $x = 1, x = \sqrt{2}, x = 2, x = \sqrt{6}, x = 2\sqrt{2}, x = \sqrt{10}, x = 2\sqrt{3}, x = \sqrt{14}$, where the function experiences a jump discontinuity.

The numerator $k^{3+}6k^{2+}11k+5$ can be rearranged into the form $(k+1)(k+2)(k+3) - 1$. Dividing this expression by $(k+3)!$ allows us to split the sum into two simpler fractions, $\frac{1}{k!} - \frac{1}{(k+3)!}$, facilitating the evaluation of the infinite series based on the expansion of $e = \sum_{k=0}^{\infty } \frac{1}{k!}$.

The first part of the series, $\sum_{k=1}^{\infty } \frac{1}{k!}$, converges to $e-1$. The second part, $\sum_{k=1}^{\infty } \frac{1}{(k+3)!}$, represents the sum of reciprocal factorials starting from $\frac{1}{4!}$, which equates to $e - 1 - \frac{1}{1!} - \frac{1}{2!} - \frac{1}{3!}$. Subtracting this second series from the first effectively cancels out the $e$ terms, leaving the calculation as $1 + \frac{1}{2} + \frac{1}{6}$, which results in $5/3$.

Because $\cos |x|$ is equivalent to $\cos x$, it is differentiable for all real numbers and does not contribute to any non-differentiable points. Consequently, the non-differentiability of $f(x)= (x^{2-}1)\left| x^{2-}a x+2 \right|+\cos |x|$ must originate from the absolute value term $\left| x^{2-}a x+2 \right|$, which typically fails to be differentiable where the expression inside equals zero. Given that $x=\alpha=2$ is one such point, substituting it into the quadratic $x^{2-}ax+2=0$ leads to $4-2a+2=0$, confirming that $a=3$.