# 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

(c)     – 5

(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

(c)     7

(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 $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(3x – 1) / x ] is

(a)    103log103

(b)    log3e

(c)    loge3

(d)    1

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(3x – 1) / x ] = loge3. [We know,  $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ (ax-1) / x = logea ]

So, option (c) is correct

6. Evaluate $\displaystyle \underset{{x\to t}}{\mathop{{\lim }}}\,$ {(x3 – t3) /  (x2 – t2)}

(a)     $\displaystyle \frac{3}{2}$

(b)      $\displaystyle \frac{1}{3}$

(c)       $\displaystyle \left( {\frac{{3t}}{2}} \right)$

(d)      1

$\displaystyle \underset{{x\to t}}{\mathop{{\lim }}}\,$ {(x3 – t3) /  (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

(c)     11

(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 +  5x. 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

(c)    10

(d)    12

$\displaystyle \underset{{x\to 4}}{\mathop{{\lim }}}\,$ {(x216) /  (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 $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {( ex – e-x) / x ]

(a)       ex

(b)      3

(c)       1

(d)      2

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

[multiplying numerator & denominator by ex].

$\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ [ {(e2x -1 ) } / ( ex. x)} = $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$ {(e2x -1 ) / x}  X (1 / ex)]

=  $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$  {(e2x -1 ) / x}  X $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$  (1 / ex)

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

So,   $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$  {(e2x -1 ) / x}  X $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$  (1 / ex

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

= 2 X $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,$   {(ex -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

(c)     2h + 11

(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)      $\displaystyle -\frac{2}{3}$

(c)       $\displaystyle \frac{1}{3}$

(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

(c)     – 1

(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

(c)     1 / 2√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

(c)    3

(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

(c)     5

(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)    $\displaystyle \frac{1}{7}$

(c)    $\displaystyle \frac{1}{5}$

(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

(c)  – 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)   $\displaystyle -\frac{1}{3}$

(c)  0

(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) –  $\displaystyle \frac{2}{9}$

(c)  $\displaystyle \frac{2}{9}$

(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

(c)     8, – 8

(d)     None of these

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

= 9a8 ($\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,$(x9-a9) / (x-a)} = n.an-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)    $\displaystyle \frac{1}{2}$

(c)    $\displaystyle \frac{1}{4}$

(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

(c)    2

(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