Question 6

   Question: $a = \operatorname{lim}_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4}$ and $b = \operatorname{lim}_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}}$, then the value of $ab^3$ is:

   Options: A. 36 B. 32 C. 25 D. 30

   Correct Answer: B

   Year: 27-Jan-2024 Shift 1

   Solution: 

    $a = \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2}}{x^4} = \frac{1}{4\sqrt{2}}$.

    $b = \lim_{x \to 0} \frac{\sin^2 x}{\sqrt{2} - \sqrt{1 + \cos x}} = 4\sqrt{2}$.

    $ab^3 = \frac{1}{4\sqrt{2}} \times (4\sqrt{2})^3 = 32$.

   Step Solution:

    1.  Solve for 'a': Rationalize the numerator of 'a' twice to eliminate the indeterminate form.

    2.  Simplify 'a': After rationalization, the expression becomes $\lim_{x \to 0} \frac{x^4}{x^4(\sqrt{1+\sqrt{1+x^4}} + \sqrt{2})(\sqrt{1+x^4} + 1)}$.

    3.  Evaluate 'a': Substituting $x=0$ gives $a = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.

    4.  Solve for 'b': Rationalize the denominator of 'b' and simplify $(1-\cos^2 x)$ to $(1-\cos x)(1+\cos x)$, allowing the $(1-\cos x)$ terms to cancel.

    5.  Final Calculation: $b = 4\sqrt{2}$. Thus, $ab^3 = \frac{1}{4\sqrt{2}} \times 128\sqrt{2} = 32$.

   Difficulty Level: Medium

   Concept Name: Rationalization of Limits

   Short cut solution: Use Taylor series expansions for $\sqrt{1+t} \approx 1 + \frac{t}{2}$ and $\cos x \approx 1 - \frac{x^2}{2}$ to quickly simplify the radicals.

Question 22

   Question: $\operatorname{lim}_{x \to \infty} \frac{(\sqrt{3x+1} + \sqrt{3x-1})^6 + (\sqrt{3x+1} - \sqrt{3x-1})^6}{(x + \sqrt{x^2-1})^6 + (x - \sqrt{x^2-1})^6} x^3$

   Options: A. is equal to 9 B. is equal to 27 C. does not exist D. is equal to $\frac{27}{2}$

   Correct Answer: B

   Year: 31-Jan-2023 Shift 2

   Solution: $\operatorname{lim}_{x \to \infty} \frac{x^3 \{ ( \sqrt{3 + 1/x} + \sqrt{3 - 1/x} )^6 + ( \sqrt{3 + 1/x} - \sqrt{3 - 1/x} )^6 \}}{x^6 \{ ( 1 + \sqrt{1 - 1/x^2} )^6 + ( 1 - \sqrt{1 - 1/x^2} )^6 \}} = \frac{(2\sqrt{3})^6 + 0}{2^6 + 0} = 3^3 = 27$.

   Step Solution:

    1.  Factor numerator: Factor out $\sqrt{x}$ from each square root in the numerator, resulting in $(\sqrt{x})^6 = x^3$.

    2.  Factor denominator: Factor out $x$ from the terms in the denominator, resulting in $x^6$.

    3.  Combine $x$ terms: The $x^3$ from the numerator and the lone $x^3$ multiplier combine to $x^6$, which cancels with the $x^6$ in the denominator.

    4.  Apply limit: As $x \to \infty$, terms like $1/x$ and $1/x^2$ tend to zero.

    5.  Calculate constants: The expression simplifies to $\frac{(\sqrt{3}+\sqrt{3})^6}{2^6} = \frac{(2\sqrt{3})^6}{2^6} = 27$.

   Difficulty Level: Medium

   Concept Name: Limits at Infinity

   Short cut solution: Recognize that as $x \to \infty$, the dominant terms are $\sqrt{3x}$ in the numerator and $x$ in the denominator. The expression essentially reduces to $\frac{(2\sqrt{3x})^6}{(2x)^6} \cdot x^3 = \frac{2^6 \cdot 27 \cdot x^3}{2^6 \cdot x^6} \cdot x^3 = 27$.

Question 43

   Question: If $\lim_{x \to 1} \frac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2$, then the value of $(a - b)$ is equal to:

   Options: (Options not explicitly provided in the source excerpt, but the solution yields a specific integer).

   Correct Answer: 11

   Year: 28-Jun-2022-Shift-2

   Solution: $\operatorname{lim}_{x \to 1} \frac{3x^2 - 4x + 1 - x^2 + 1}{2x^3 - 7x^2 + ax + b} = \operatorname{lim}_{x \to 1} \frac{2(x-1)^2}{2x^3 - 7x^2 + ax + b} = -2$. Solving for constants gives $a-b=11$.

   Step Solution:

    1.  Indeterminate Form: At $x=1$, the numerator is $0$; for the limit to be finite ($-2$), the denominator must also be $0$, so $2-7+a+b=0 \implies a+b=5$.

    2.  Simplify Numerator: Using $\sin \theta \approx \theta$ for small $\theta$, the numerator becomes $(3x^2-4x+1) - (x^2-1) = 2x^2 - 4x + 2 = 2(x-1)^2$.

    3.  Identify Factor: For the limit to exist and be non-zero, the denominator must have $(x-1)^2$ as a factor.

    4.  Find Constants: Let the denominator be $(x-1)^2(2x-3)$ to match the leading coefficient and the required limit value of $-2$.

    5.  Solve for a and b: Expanding $(x-1)^2(2x-3)$ gives $2x^3 - 7x^2 + 8x - 3$. Thus $a=8$ and $b=-3$, so $a-b = 8 - (-3) = 11$.

   Difficulty Level: Hard

   Concept Name: Indeterminate Forms / Standard Limits

   Short cut solution: Use L'Hopital's rule twice. Since the numerator's second derivative at $x=1$ is $4$, the denominator's second derivative must be $-2$ at $x=1$ to satisfy the limit $-2$ (i.e., $4 / f''(1) = -2$).

Question 62

   Question: Let $k$ and $m$ be positive real numbers such that the function $f(x) = \begin{cases} 3x^2 + k\sqrt{x+1} & 0 < x < 1 \\ mx^2 + k^2 & x \ge 1 \end{cases}$ is differentiable for all $x > 0$. Then $\frac{8f'(8)}{f'(1/8)}$ is equal to.

   Options: (The source lists this as a numerical entry question; the correct value is 309).

   Correct Answer: 309.

   Year: 8-Apr-2023 shift 2.

   Solution: Since the function is differentiable at $x=1$, it must be continuous and have equal left and right derivatives at that point. Solving $f(1^-) = f(1^+)$ and $f'(1^-) = f'(1^+)$ allows us to find $k = \frac{7}{4\sqrt{2}}$ and $m = \frac{103}{32}$. Substituting these into the derivative ratio yields 309.

   Step Solution:

    1.  Continuity at $x=1$: Equate $f(1^-)$ and $f(1^+)$: $3 + k\sqrt{2} = m + k^2$.

    2.  Differentiability at $x=1$: Equate derivatives: $f'(x) = 6x + \frac{k}{2\sqrt{x+1}}$ (for $x<1$) and $2mx$ (for $x>1$). Thus, $6 + \frac{k}{2\sqrt{2}} = 2m$.

    3.  Solve for $k$ and $m$: Substitute $m = 3 + \frac{k}{4\sqrt{2}}$ into the continuity equation to find $k = \frac{7}{4\sqrt{2}}$ and then $m = \frac{103}{32}$.

    4.  Calculate $f'(8)$ and $f'(1/8)$: $f'(8) = 2(8)m = 16(\frac{103}{32}) = \frac{103}{2}$. $f'(1/8) = 6(1/8) + \frac{k}{2\sqrt{9/8}} = \frac{3}{4} + \frac{7}{12} = \frac{4}{3}$.

    5.  Final Ratio: $\frac{8 \times (103/2)}{4/3} = \frac{412}{4/3} = 309$.

   Difficulty Level: Hard

   Concept Name: Differentiability of Piecewise Functions

   Short cut solution: Quickly equate the derivatives at the boundary to find a relationship between $m$ and $k$, then use the continuity boundary condition to solve for the specific values.

 Question 75

   Question: If $y = \tan^{-1}(\sec x^3 - \tan x^3)$, $\frac{\pi}{2} < x^3 < \frac{3\pi}{2}$, then:

   Options: 

       A. $xy' + 2y' = 0$

       B. $x^2y' - 6y + \frac{3\pi}{2} = 0$

       C. $x^2y' - 6y + 3\pi = 0$

       D. $xy' - 4y' = 0$

   Correct Answer: B.

   Year: 24-Jun-2022-Shift-2.

   Solution: Let $x^3 = \theta$. Simplify the expression $\sec \theta - \tan \theta = \tan(\frac{\pi}{4} - \frac{\theta}{2})$. Thus $y = \frac{\pi}{4} - \frac{x^3}{2}$. Differentiating gives $y' = -\frac{3x^2}{2}$, which satisfies the relation in option B.

   Step Solution:

    1.  Trigonometric Simplification: Use $\sec \theta - \tan \theta = \frac{1-\sin \theta}{\cos \theta} = \frac{(\cos \theta/2 - \sin \theta/2)^2}{\cos^2 \theta/2 - \sin^2 \theta/2} = \tan(\frac{\pi}{4} - \frac{\theta}{2})$.

    2.  Simplify $y$: $y = \tan^{-1}(\tan(\frac{\pi}{4} - \frac{x^3}{2}))$. In the given range, $y = \frac{\pi}{4} - \frac{x^3}{2}$.

    3.  Differentiate: $\frac{dy}{dx} = y' = -\frac{3x^2}{2}$.

    4.  Manipulate for Options: Multiply by $x$: $x y' = -\frac{3x^3}{2}$. Since $x^3 = \frac{\pi}{2} - 2y$, then $x y' = -\frac{3}{2}(\frac{\pi}{2} - 2y) = -\frac{3\pi}{4} + 3y$.

    5.  Final Form: $2x y' - 6y + \frac{3\pi}{2} = 0$. (Note: Source option B is the closest match, though written as $x^2 y'$, likely referring to the relation $2xy'$).

   Difficulty Level: Medium

   Concept Name: Inverse Trigonometric Substitution

   Short cut solution: Recognize the standard half-angle identity $\sec \theta - \tan \theta = \tan(\frac{\pi}{4} - \frac{\theta}{2})$ to avoid heavy differentiation of the inverse tangent function.

 Question 78

   Question: If $\lim_{x \to \infty} (\sqrt{x^2 - x + 1} - ax) = b$, then the ordered pair $(a, b)$ is:

   Options: 

       A. $(1, \frac{1}{2})$

       B. $(1, -\frac{1}{2})$

       C. $(-1, \frac{1}{2})$

       D. $(-1, -\frac{1}{2})$

   Correct Answer: B.

   Year: 2021, 27 Aug. Shift-2.

   Solution: For the limit to exist as $x \to \infty$, the coefficient of the leading term must be zero, implying $a=1$. Rationalizing the expression gives $\lim_{x \to \infty} \frac{-x+1}{\sqrt{x^2-x+1}+x}$, which evaluates to $b = -1/2$.

   Step Solution:

    1.  Determine 'a': For the limit to be finite as $x \to \infty$, $a$ must be positive; specifically $a=1$ to cancel the $x^2$ term during rationalization.

    2.  Rationalize: Multiply and divide by the conjugate: $\lim_{x \to \infty} \frac{(\sqrt{x^2-x+1}-x)(\sqrt{x^2-x+1}+x)}{\sqrt{x^2-x+1}+x}$.

    3.  Simplify Numerator: $(x^2 - x + 1) - x^2 = -x + 1$.

    4.  Divide by $x$: Evaluate $\lim_{x \to \infty} \frac{-1 + 1/x}{\sqrt{1 - 1/x + 1/x^2} + 1}$.

    5.  Evaluate Limit: The limit is $\frac{-1}{1+1} = -1/2$, so $b = -1/2$.

   Difficulty Level: Medium

   Concept Name: Limits at Infinity and Rationalization

   Short cut solution: Use the binomial expansion for large $x$: $\sqrt{x^2-x+1} = x(1 - \frac{1}{x} + \frac{1}{x^2})^{1/2} \approx x(1 - \frac{1}{2x}) = x - \frac{1}{2}$. Then $(x - \frac{1}{2}) - ax = b$ immediately gives $a=1$ and $b=-1/2$.

Question 86

   Question: $\operatorname { l i m } _ { \mathbf { x } \to 2 } { \frac { 3 ^ { \mathbf { x } } + 3 ^ { 3 - \mathbf { x } } - 1 2 } { 3 ^ { - { \frac { \mathbf { x } } { 2 } } } - 3 ^ { 1 - \mathbf { x } } } }$

   Options: (The source lists this as a numerical entry question; no A, B, C, D options are provided).

   Correct Answer: 36.

   Year: Jan. 7, 2020 (Shift I).

   Solution: Let $3^x = t^2$. Then $\lim_{t \to 3} \frac{t^2 + 27/t^2 - 12}{1/t - 3/t^2} = \lim_{t \to 3} \frac{t^4 - 12t^2 + 27}{t - 3} = \lim_{t \to 3} \frac{(t^2-9)(t^2-3)}{t-3} = \lim_{t \to 3} (t+3)(t^2-3) = (6)(6) = 36$.

   Step Solution:

    1.  Substitution: Let $3^x = t^2$. As $x \to 2$, $t^2 \to 9$, so $t \to 3$.

    2.  Rewrite Expression: The limit becomes $\lim_{t \to 3} \frac{t^2 + 27/t^2 - 12}{1/t - 3/t^2}$.

    3.  Clear Denominators: Multiply the numerator and denominator by $t^2$ to get $\lim_{t \to 3} \frac{t^4 - 12t^2 + 27}{t - 3}$.

    4.  Factorize: Recognize $t^4 - 12t^2 + 27$ as a quadratic in $t^2$: $(t^2 - 9)(t^2 - 3) = (t - 3)(t + 3)(t^2 - 3)$.

    5.  Evaluate: Cancel $(t - 3)$ and substitute $t = 3$: $(3 + 3)(3^2 - 3) = 6 \times 6 = 36$.

   Difficulty Level: Medium

   Concept Name: Substitution in Limits / Algebraic Manipulation

   Short cut solution: Apply L'Hopital's Rule directly to the original form, though the differentiation of the exponents might be slightly more tedious than substitution.

 Question 96

   Question: $\operatorname { l i m } _ { \mathbf { x } \to 0 } { \frac { \mathbf { x } { \big ( } \mathbf { e } ^ { { \big ( } { \sqrt { 1 + \mathbf { x } ^ { 2 } + \mathbf { x } ^ { 4 } } } - 1 { \big ) } / \mathbf { x } } - 1 { \big ) } } { \sqrt { 1 + \mathbf { x } ^ { 2 } + \mathbf { x } ^ { 4 } } - 1 } }$

   Options: A. is equal to $\sqrt{e}$; B. is equal to 1; C. is equal to 0; D. does not exist.

   Correct Answer: B.

   Year: Sep. 05, 2020 (Shift II).

   Solution: Let $t = \frac{\sqrt{1+x^2+x^4}-1}{x}$. As $x \to 0$, $t \to 0$ (via rationalization). The expression simplifies to $\lim_{t \to 0} \frac{e^t - 1}{t} = 1$.

   Step Solution:

    1.  Identify the variable: Let $u = \frac{\sqrt{1+x^2+x^4}-1}{x}$.

    2.  Find limit of $u$: Rationalize the numerator of $u$ to show $\lim_{x \to 0} u = \lim_{x \to 0} \frac{x^2+x^4}{x(\sqrt{1+x^2+x^4}+1)} = \lim_{x \to 0} \frac{x+x^3}{2} = 0$.

    3.  Restructure the limit: Notice the denominator is $x \cdot u$. The original expression is $\frac{x(e^u - 1)}{x \cdot u}$.

    4.  Simplify: The $x$ terms cancel, leaving $\frac{e^u - 1}{u}$.

    5.  Apply standard limit: Since $u \to 0$ as $x \to 0$, $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$.

   Difficulty Level: Medium

   Concept Name: Standard Exponential Limit ($\lim_{t \to 0} \frac{e^t-1}{t}=1$)

   Short cut solution: Use Taylor expansion for the exponent: $\sqrt{1+x^2+x^4}-1 \approx (1 + \frac{x^2}{2}) - 1 = \frac{x^2}{2}$. The exponent becomes $\frac{x^2/2}{x} = \frac{x}{2}$. The limit then looks like $\frac{x(e^{x/2}-1)}{x^2/2} \approx \frac{x(x/2)}{x^2/2} = 1$.

 Question 102

   Question: $\operatorname { l i m } _ { \mathbf { y } \to 0 } { \frac { \sqrt { 1 + { \sqrt { 1 + { \mathbf { y } } ^ { 4 } } } } - { \sqrt { 2 } } } { \mathbf { y } ^ { 4 } } }$

   Options: A. exists and equals $\frac{1}{4\sqrt{2}}$; B. exists and equals $\frac{1}{2\sqrt{2}(\sqrt{2}+1)}$; C. exists and equals $\frac{1}{2\sqrt{2}}$; D. does not exist.

   Correct Answer: A.

   Year: Jan. 9, 2019 (Shift I).

   Solution: Rationalize the numerator twice. $L = \lim_{y \to 0} \frac{(1+\sqrt{1+y^4})-2}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})} = \lim_{y \to 0} \frac{\sqrt{1+y^4}-1}{y^4(\dots)} = \frac{1}{4\sqrt{2}}$.

   Step Solution:

    1.  First Rationalization: Multiply by $\frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$ to get $\lim_{y \to 0} \frac{1+\sqrt{1+y^4}-2}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})}$.

    2.  Simplify Numerator: This yields $\frac{\sqrt{1+y^4}-1}{y^4(\dots)}$.

    3.  Second Rationalization: Multiply by $\frac{\sqrt{1+y^4}+1}{\sqrt{1+y^4}+1}$ to get $\lim_{y \to 0} \frac{1+y^4-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$.

    4.  Cancel $y^4$: The $y^4$ in the numerator and denominator cancel.

    5.  Evaluate: Substitute $y=0$: $\frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.

   Difficulty Level: Medium

   Concept Name: Double Rationalization

   Short cut solution: Use Taylor expansion twice: $\sqrt{1+y^4} \approx 1 + \frac{y^4}{2}$. Then $\sqrt{1+(1+y^4/2)} = \sqrt{2+y^4/2} = \sqrt{2}\sqrt{1+y^4/4} \approx \sqrt{2}(1 + \frac{y^4}{8}) = \sqrt{2} + \frac{\sqrt{2}y^4}{8}$. The limit becomes $\frac{(\sqrt{2} + \frac{\sqrt{2}y^4}{8}) - \sqrt{2}}{y^4} = \frac{\sqrt{2}}{8} = \frac{1}{4\sqrt{2}}$.

 Question 105

   Question: $\operatorname { l i m } _ { \mathbf { x } \to 0 } { \frac { \mathbf { x } + 2 \sin \mathbf { x } } { \sqrt { \mathbf { x } ^ { 2 } + 2 \sin \mathbf { x } + 1 } - { \sqrt { \sin ^ { 2 } \mathbf { x } - \mathbf { x } + 1 } } } }$

   Options: A. 6, B. 2, C. 3, D. 1

   Correct Answer: B

   Year: April 12, 2019 (II)

   Solution: Rationalize the denominator by multiplying by the conjugate:

    $\operatorname { l i m } _ { \mathbf { x } \to 0 } \frac { ( \mathbf { x } + 2 \sin \mathbf { x } ) [ \sqrt { \mathbf { x } ^ { 2 } + 2 \sin \mathbf { x } + 1 } + \sqrt { \sin ^ { 2 } \mathbf { x } - \mathbf { x } + 1 } ] } { ( \mathbf { x } ^ { 2 } - \sin ^ { 2 } \mathbf { x } ) + ( \mathbf { x } + 2 \sin \mathbf { x } ) } = \frac{3 \times 2}{3} = 2$.

   Step Solution:

    1.  Identify Form: At $x=0$, the limit is in the $\frac{0}{0}$ form.

    2.  Rationalize Denominator: Multiply numerator and denominator by the conjugate $(\sqrt { \mathbf { x } ^ { 2 } + 2 \sin \mathbf { x } + 1 } + { \sqrt { \sin ^ { 2 } \mathbf { x } - \mathbf { x } + 1 } })$.

    3.  Simplify Denominator: The denominator becomes $(x^2 + 2\sin x + 1) - (\sin^2 x - x + 1) = x^2 - \sin^2 x + x + 2\sin x$.

    4.  Factor $x$: Divide both numerator and denominator by $x$. The numerator becomes $(1 + \frac{2\sin x}{x}) \times (\text{conjugate terms})$ and denominator becomes $(x - \frac{\sin^2 x}{x} + 1 + \frac{2\sin x}{x})$.

    5.  Apply Standard Limits: Since $\lim_{x \to 0} \frac{\sin x}{x} = 1$, the expression evaluates to $\frac{(1+2) \times (1+1)}{(0-0+1+2)} = \frac{3 \times 2}{3} = 2$.

   Difficulty Level: Medium

   Concept Name: Rationalization and Standard Limits

   Short cut solution: Use expansion: $\sin x \approx x$. Numerator $\approx x + 2x = 3x$. Denominator $\approx \sqrt{x^2+2x+1} - \sqrt{x^2-x+1} \approx (1+2x)^{1/2} - (1-x)^{1/2} \approx (1+x) - (1-x/2) = 1.5x$. Limit is $3x/1.5x = 2$.

 Question 113

   Question: $\operatorname { l i m } _ { \mathbf { x } \to 0 } \frac{x}{9 - ( 27 + \mathbf { x } ) ^ { \frac { 1 } { 3 } } - 3}$ (Transcription note: The source shows a fragmented numerator, but the solution focuses on the $27+x$ radical).

   Options: A. $-\frac{1}{3}$, B. $\frac{1}{6}$, C. $-\frac{1}{6}$, D. $\frac{1}{3}$

   Correct Answer: C

   Year: Online April 16, 2018

   Solution: Let $L = \operatorname { l i m } _ { x \to 0 } \frac{\dots}{9 - ( 27 + x )^{1/3}}$. Using L'Hospital rule, $L = \frac{1/3 \times (27)^{-2/3}}{-2/3 \times 27^{-1/3}} = -1/6$. (Note: Source transcription contains some notation errors).

   Step Solution:

    1.  Identify Form: The limit is in the indeterminate $\frac{0}{0}$ form.

    2.  Apply L'Hospital Rule: Differentiate the numerator and the denominator with respect to $x$.

    3.  Derivative of Denominator: The derivative of $9 - (27+x)^{1/3}$ is $-\frac{1}{3}(27+x)^{-2/3}$.

    4.  Substitute $x=0$: Evaluate the resulting expression at $x=0$.

    5.  Final Calculation: Following the source logic, the simplified constant ratio results in $-\frac{1}{6}$.

   Difficulty Level: Easy

   Concept Name: L'Hospital's Rule

   Short cut solution: Use Binomial Expansion: $(27+x)^{1/3} = 3(1 + \frac{x}{27})^{1/3} \approx 3(1 + \frac{x}{81}) = 3 + \frac{x}{27}$. Substitute and simplify the expression to find the limit quickly.

 Question 116

   Question: $\operatorname { l i m } _ { \mathrm { x } \to 3 } \frac { { \sqrt { 3 \mathrm { x } } } - 3 } { \sqrt { 2 \mathrm { x } - 4 } - { \sqrt { 2 } } }$

   Options: A. $\sqrt{3}$, B. 1, C. $\frac{\sqrt{3}}{2}$, D. $\frac{1}{2\sqrt{2}}$

   Correct Answer: B (Note: Solution in calculates $1/\sqrt{2}$, which is closer to option D).

   Year: Online April 8, 2017

   Solution: Rationalize the expression: $\operatorname { l i m } _ { \mathrm { x } \to 3 } \frac { 3 ( \mathbf { x } - 3 ) } { ( 2 \mathbf { x } - 6 ) } \times \frac { \sqrt { 2 \mathbf { x } - 4 } + \sqrt { 2 } } { \sqrt { 3 \mathbf { x } } + 3 } = \frac { 3 } { 2 } \times \frac { 2 \sqrt { 2 } } { 6 } = \frac { 1 } { \sqrt { 2 } }$.

   Step Solution:

    1.  Rationalize Numerator: Multiply and divide by $(\sqrt{3x} + 3)$ to get $(3x - 9)$ in the numerator.

    2.  Rationalize Denominator: Multiply and divide by $(\sqrt{2x - 4} + \sqrt{2})$ to get $(2x - 6)$ in the denominator.

    3.  Factorize: Rewrite $(3x-9)$ as $3(x-3)$ and $(2x-6)$ as $2(x-3)$.

    4.  Cancel Terms: Cancel the common $(x-3)$ term from the numerator and denominator.

    5.  Substitute $x=3$: The limit becomes $\frac{3}{2} \times \frac{\sqrt{2} + \sqrt{2}}{3+3} = \frac{3}{2} \times \frac{2\sqrt{2}}{6} = \frac{1}{\sqrt{2}}$.

   Difficulty Level: Medium

   Concept Name: Rationalization of Limits

   Short cut solution: Use L'Hospital's Rule: Numerator derivative is $\frac{3}{2\sqrt{3x}} \to \frac{3}{6} = 1/2$. Denominator derivative is $\frac{2}{2\sqrt{2x-4}} \to \frac{2}{2\sqrt{2}} = 1/\sqrt{2}$. Limit $= (1/2) / (1/\sqrt{2}) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.