Limits & Continuity MCQ -

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Limits & Continuity MCQ

1.Value of $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ {(2x²+9x+9) / (2x²+7x+3)} is

(a) $\displaystyle \frac{3}{5}$

(b) $\displaystyle \frac{5}{3}$

(c) $\displaystyle \frac{3}{4}$

(d) $\displaystyle \frac{1}{6}$

$\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ {(2x²+9x+9) / (2x²+7x+3) }

= { $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x²+9x+9) } / { $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x²+7x+3) }

= { $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x²+6x+3x+9) } / { $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x²+6x+x+3) }

= { $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x(x+3)+3(x+3)} / $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ {2x(x+3)+1(x+3)}

= $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x+3)(x+3))} / { $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x+1)(x+3)}

= $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x+3) / $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ (2x+1) = $\displaystyle \underset{{x\to -3}}{\mathop{{\lim }}}\,$ = $\displaystyle \frac{{-3}}{{-5}}$ = $\displaystyle \frac{3}{5}$ .
So, option (a) is correct

2. Value of $\displaystyle \underset{{n\to 4}}{\mathop{{\lim }}}\,$ {(x²– 3x-4) / (x²– 2x-8)} is

(a) $\displaystyle \frac{1}{6}$

(b) $\displaystyle \frac{5}{6}$

(c) $\displaystyle -\frac{5}{6}$

(d) $\displaystyle -\frac{3}{4}$

$\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ {(x²– 3x-4) / (x²-2x-8)}

= $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ {(x² + x – 4x -4) / (x² +2x – 4x -8)}

= $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ [{ x (x+1) – 4 (x+1)} / {x (x+2) – 4 (x+2)}}]

= $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ { (x+1) (x– 4) / (x+2) (x– 4)} = $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ {(x+1) / (x+2)

= $\displaystyle \frac{{4+1}}{{4+2}}=\frac{5}{6}$ .
So, option (b) is correct

3. Find Value of $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x²– 4) / {√(x+2) – √(3x-2)}

(a) – 3

(b) – 4

(d) – 8.

$\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x+2)( x-2) / {√(x+2) – √(3x-2)}

= $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x+2)( x-2) {√(x+2) + √(3x-2)} / {√(x+2) – √(3x-2)} {√(x+2) + √(3x-2)}

[multiplying numerator and denominator by {√(x+2) + √(3x-2)}]

$\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x+2)( x-2) {√(x+2) + √(3x-2)} / {√(x+2)² – √(3x-2)²}

= $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x+2)( x-2) {√(x+2) +√(3x-2)} / {(x+2) – (3x-2)}]

= $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x+2)( x-2) {√(x+2) +√(3x-2)} / {(x+2- 3x+2)}]

= $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x+2)( x-2) {√(x+2) +√(3x-2)} / (4-2x)]

= $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x+2)( x-2) {√(x+2) +√(3x-2)} / -2(x-2)]

= $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x+2){√(x+2) +√(3x-2)} / -2]

= $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(2+2){√(2+2) +√{(3 X2) -2)} / -2]

= 4 ( √4 + √ 4) / -2 = (4 (2+2)} / – 2 = 16 / -2 = – 8

So, option (d) is correct

4. Value of $\displaystyle \underset{{x\to 2}}{\mathop{{\lim }}}\,$ [(x²– 4) / (x+2)] is

(a) 4

(b) – 4

(d) – 6

$\displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,$ [(x²– 4) / (x+2)] = $\displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,$ [{(x + 2) (x – 2)} / (x+2)]

= $\displaystyle \underset{{x\to -2}}{\mathop{{\lim }}}\,$ (x – 2) = -2 – 2 = – 4. So, option (b) is correct

5. Value of $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {₍₃^x_{– 1)} / x ] is

(a) 10³log₁₀3

(b) log₃e

(d) 1

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {₍₃^x_{– 1)} / x ] = log_e³. [We know, $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (a^x-1) / x = log_e^a ]

So, option (c) is correct

6. Evaluate $\displaystyle \underset{{x\to t}}{\mathop{{\lim }}}\,$ {(_x³_{– t}³) / (x² – t²)}

(a) $\displaystyle \frac{3}{2}$

(b) $\displaystyle \frac{1}{3}$

(d) 1

$\displaystyle \underset{{x\to t}}{\mathop{{\lim }}}\,$ {(_x³_{– t}³) / (x² – t²)} = $\displaystyle \underset{{x\to t}}{\mathop{{\lim }}}\,$ [ {(x – t) (x² + x t + t²)} / {(x – t) (x + t)}

= $\displaystyle \underset{{x\to t}}{\mathop{{\lim }}}\,$ [ {(x² + x t + t²)} / (x + t)}

= (t² + t² + t²) / (t + t) = (3 t²) / 2t = $\displaystyle \frac{{3t}}{2}$

So, option (c) is correct

7. Evaluate $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(4x⁴ + 5x³. 7x² + 6x) / (5x⁵ + 7x² + x) ]

(a) 6

(b) 9

(d) 8

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(4x⁴ + 5x³. 7x² + 6x) / (5x⁵ + 7x² + x)]

= $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(x{(4x³ + 5x³. 7x + 6)} / x(5x⁴ + 7x + 1)}

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {{(4x³ + 5x³7.x + 6)} / (5x⁴ + 7x + 1)}

= $\displaystyle \frac{{0+0+6}}{{0+0+1}}=\frac{6}{1}$ = 6

So, option (a) is correct

8. Evaluate $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ {(_x²_{– 16}) / (x- 4)}

(a) 8

(b) – 8

(d) 12

$\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ {(_x²_–16) / (x- 4)} = $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ {(x – 4) (x+4) / (x- 4)}

= $\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ (x+4) = 4+4=8

So, option (a) is correct

9. Evaluate $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {( e^x – e^-x) / x ]

(a) e^x

(b) 3

(d) 2

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {( e^x – e^-x) / x ] = $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ [{ e^x ( e^x – e^-x) } / ( e^x. x) ]

[multiplying numerator & denominator by e^x].

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ [ {(e^2x -1 ) } / ( e^x. x)} = $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(e^2x -1 ) / x} X (1 / e^x)]

= $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(e^2x -1 ) / x} X $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (1 / e^x)

Let 2x=z, so x=z / 2. (z 0, when x 0).

So, $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(e^2x -1 ) / x} X $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (1 / e^x)

= $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(e^z -1 ) / (z/2) } X 1 / 1 (as $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ e^x =1)

= 2 X $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(e^x -1 ) / x = 2 X 1 X 1 =2

So, option (d) is correct

10. If f(x) = $\displaystyle 2{{x}^{2}}-x+2$ then evaluate {f(4 + h) – f(4)} / h

(a) 4h

(b) 15 + 2h

(d) 3h

f(x) = 2x² – x + 2

So , f(4+h) = 2. (4+h)² – (4+h) + 2

= 2 (16 + 8h + h²) – 4 – h + 2

= 32 + 16h + 2h² – h – 2 = 30 + 15h + 2h²

Again, f(4) = 2. (4)² – 4 + 2 = 2. 16 – 4 + 2 = 32 – 2 = 30

So, f(4+h) – f(4) = (30 + 15h + 2h²) – 30

= 15h + 2h² = h (15 + 2h)

So, {f(4+h) – f(4)} / h = {h (15 + 2h)} / h = (15 + 2h)

So, option (b) is correct

11. Evaluate $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ {(_2x²+_{5x – 7}) / (3x² – 7x + 9)}

(a) 0

(b) $\displaystyle -\frac{2}{3}$

(d) $\displaystyle \frac{2}{3}$

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ {(_2x²+_{5x – 7}) / (3x² – 7x + 9)}

= $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ {((2 + –7 / x²) / (3 – 7 / x² + 9 / x²)

[Dividing numerator & denominator by x²]

= $\displaystyle \frac{{2+0-0}}{{3-0+0}}$ = . [When $\displaystyle x\to \infty$ , $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\frac{1}{x}$ = 0 (also $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ 1 / x² =0)]

So, option (d) is correct

12. Evaluate $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ √((1+x²) – √((1-x²)} / x]

(a) 3

(b) 0

(d) 4

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ √((1+x²) – √((1-x²)}/ x]

= $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ √{(1+x²) – √((1-x²)} X √{(1+x²) + √((1-x²)} / { x√{(1+x²) + √((1-x²)}]

(Multiplying numerator & denominator by √{(1+x²) + √((1-x²)}]

= $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (1 + x² – (1-x²)} / {x√{(1+x²) + √((1-x²)}]

= $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ 2x² / {x√{(1+x²) + √((1-x²)}]

= $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ [(2x ) / {√{(1+x²) + √((1-x²)}]

= {2 X 0 / (√1 + √1)} = $\displaystyle \frac{0}{{1+1}}$ = $\displaystyle \frac{0}{2}$ = 0

So, option (b) is correct

13. Evaluate $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ √((x+h) – √x} / h, if h>0

(a) 3√x

(b) √x

(d) x

$\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ [{√((x+h) – √x} / h]

= $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ √((x+h) – √x} {√((x+h) + √x} / √((x+h) + √x}

(Multiplying numerator & denominator by {√((x+h) + √x}

= $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ {√(x+h)² – (√x)²} / $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ √((x+h) + √x}

= $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ {(x + h – x)} / $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ √(x+h) + √x}]

$\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ h / $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ √{(x+h) + √x}}]

= $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ 1 / √(x+h) + √x}]

= 1/ √(x+0) + √x}] = 1 / (2√x)

So, option (c) is correct

14. Evaluate $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,$ (x²+1) / (x+1)}

(a) 2

(b) 1

(d) 4

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,$ (x²+1) / (x+1)} = (1² + 1) / (1+1) = = 1

So, option (b) is correct

15. Evaluate $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,$ {(x² – 1) / (x-1)}

(a) 1

(b) 2

(d) 7

$\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,$ (x² – 1) / (x-1)} = $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,$ (x – 1) (x+1)} / (x-1) }

= $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,$ (x+1) = 1+1 =2

So, option (b) is correct

16. Evaluate $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ [{1- $\displaystyle \sqrt{{}}$ (1-x)} / x]

(a) $\displaystyle \frac{1}{9}$

(b) $\displaystyle \frac{1}{7}$

(d) $\displaystyle \frac{1}{2}$

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (1-x)} / x}

= $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ [(1-x) } X $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (1-x) } / {x $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (1-x) } ]

Multiply numerator & denominator by (1 + $\displaystyle \sqrt{{}}$ 1 – x)

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(1² – $\displaystyle \sqrt{{}}$ (1-x²)} / (x(1 + $\displaystyle \sqrt{{}}$ 1 – x)} = $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (1-1+x)

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (1-1+x) / (x(1 + $\displaystyle \sqrt{{}}$ (1 – x) = $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {x / (x(1 + $\displaystyle \sqrt{{}}$ (1 – x)}

= 1 / (1+ $\displaystyle \sqrt{{}}$ 1 – 0) = $\displaystyle \frac{1}{{1+1}}$ = 2

So, option (d) is correct

17. If f(x) = x², evaluate, $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ $\displaystyle \left[ {\frac{{f(x+h)-f(x)}}{h}} \right]$

(a) 2x

(b) x

(d) 3

f(x) = x², So, f (x + h) = (x + h)² = x² + 2xh + h².

$\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ [{f(x + h) – f(x)} / h]= $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ [{(x² + 2xh + h²) – x² / h}]

= $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ [{(2xh + h²) / h}]= $\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ $\displaystyle \left[ {\left{ {\frac{{h(2x+h)}}{h}} \right}} \right]$

$\displaystyle \underset{{h\to 0}}{\mathop{{\lim }}}\,$ (2x + h) = 2x.

So, option (a) is correct

18. Evaluate $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ (6-5x²) / (4x+15x²)}

(a) $\displaystyle \frac{1}{3}$

(b) $\displaystyle -\frac{1}{3}$

(d) Limit does not exist

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ (6-5x²) / (4x+15x²)}

= $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ (6 / x² – 5) / ( $\displaystyle \frac{4}{x}$ + 15)} ((Dividing numerator & denominator by x²)

= $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ (6 X 1/ x² – 5) / (4 X $\displaystyle \frac{1}{x}$ +15)}

= $\displaystyle \frac{{\left( {6\times 0} \right)-5}}{{\left( {4\times 0} \right)+15}}$

= $\displaystyle \frac{{0-5}}{{0+15}}$ = $\displaystyle -\frac{5}{{15}}=-\frac{1}{3}$

So, option (b) is correct

19. Evaluate $\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,$ {(x²+4x+3) / (x² – 7x -8)}

(a) $\displaystyle \frac{1}{9}$

(b) – $\displaystyle \frac{2}{9}$

(d) $\displaystyle -\frac{3}{8}$

$\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,$ {(_x²+_{4x +3}) / (x² – 7x -8)}

= $\displaystyle \underset{{x\to -1}}{\mathop{{\lim }}}\,$ {(x²+x+3x+3) / (x² + x – 8x -8)}

= $\displaystyle \frac{{-1+3}}{{-1-8}}=\frac{2}{{-9}}=-\frac{2}{9}$

So, option (b) is correct

20. If $\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,$ (x⁹-a⁹) / (x-a)} =9, find the value of a.

(a) 9, -9

(b) 1, -1

(d) None of these

$\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,$ (x⁹-a⁹) / (x-a)} =9.a^9-1.

= 9a⁸ ( $\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,$ (x⁹-a⁹) / (x-a)} = n.a^n-1)

So, 9a⁸ =9, or a⁸ =1, or a⁸ = 1,

So, option (b) is correct

22. Evaluate $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ {(_2x²+_{7x +5}) / (4x² +3x + 1)}

(a) $\displaystyle -\frac{1}{2}$

(b) $\displaystyle \frac{1}{2}$

(d) 3

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ {(2x²+7x+5) / (4x² +3x + 1)}

= $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ {(2+ $\displaystyle \frac{7}{x}$ +5/ x²) / (4 + $\displaystyle \frac{3}{x}$ + 1 / x²)}

Multiply numerator & denominator by 1 / x².

$\displaystyle \frac{{2+0+0}}{{4+0+0}}$ (as $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ 1 / x² =0) = $\displaystyle \frac{2}{4}$ = $\displaystyle \frac{1}{2}$

So, option (b) is correct

23. $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ (3 / x² + 2)

(a) 1

(b) 9

(d) 7

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ (3 / x² + 2) = $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ (3 X 1 / x² + 2)= 0+2 = 2 (as $\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,$ 1 / x²= 0)

So, option (c) is correct

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