in exercises 49 - 56, evaluate in terms of the constant a.
49. $lim_{x
ightarrow0}(2a + x)$
50. $lim_{h
ightarrow - 2}(4ah+7a)$
51. $lim_{t
ightarrow - 1}(4t - 2at + 3a)$
52. $lim_{x
ightarrow a}\frac{(x + a)^{2}-4x^{2}}{x - a}$
53. $lim_{x
ightarrow a}\frac{sqrt{x}-sqrt{a}}{x - a}$
54. $lim_{h
ightarrow0}\frac{sqrt{a + 2h}-sqrt{a}}{h}$
55. $lim_{x
ightarrow0}\frac{(x + a)^{3}-a^{3}}{x}$
56. $lim_{h
ightarrow a}\frac{\frac{1}{h}-\frac{1}{a}}{h - a}$
57. evaluate $lim_{h
ightarrow0}\frac{sqrt4{1 + h}-1}{h}$. hint: set $x=sqrt4{1 + h}$, express h as a function and rewrite as a limit as $x
ightarrow1$.
58. evaluate $lim_{h
ightarrow0}\frac{sqrt3{1 + h}-1}{sqrt2{1 + h}-1}$. hint: set $x=sqrt6{1 + h}$, express h as a function of and rewrite as a limit as $x
ightarrow1$

Question

Accepted Answer

# Explanation for 49:
## Step1: Direct substitution x=0
$\lim_{x 	o 0} (2a + x) = 2a + 0$
# Answer for 49:
$2a$

# Explanation for 50:
## Step1: Substitute h=-2
$\lim_{h 	o -2} (4ah + 7a) = 4a(-2) + 7a$
## Step2: Simplify expression
$-8a + 7a = -a$
# Answer for 50:
$-a$

# Explanation for 51:
## Step1: Substitute t=-1
$\lim_{t 	o -1} (4t - 2at + 3a) = 4(-1) - 2a(-1) + 3a$
## Step2: Simplify terms
$-4 + 2a + 3a = -4 + 5a$
# Answer for 51:
$5a - 4$

# Explanation for 52:
## Step1: Expand numerator
$(x+a)^2 - 4x^2 = x^2 + 2ax + a^2 - 4x^2 = -3x^2 + 2ax + a^2$
## Step2: Factor numerator
$-3x^2 + 2ax + a^2 = -(3x + a)(x - a)$
## Step3: Cancel (x - a) and substitute x=a
$\lim_{x 	o a} \frac{-(3x + a)(x - a)}{x - a} = -(3a + a) = -4a$
# Answer for 52:
$-4a$

# Explanation for 53:
## Step1: Rationalize numerator
$\frac{\sqrt{x} - \sqrt{a}}{x - a} = \frac{(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a})}{(x - a)(\sqrt{x} + \sqrt{a})} = \frac{x - a}{(x - a)(\sqrt{x} + \sqrt{a})}$
## Step2: Cancel (x - a) and substitute x=a
$\lim_{x 	o a} \frac{1}{\sqrt{x} + \sqrt{a}} = \frac{1}{2\sqrt{a}}$
# Answer for 53:
$\frac{1}{2\sqrt{a}}$

# Explanation for 54:
## Step1: Rationalize numerator
$\frac{\sqrt{a + 2h} - \sqrt{a}}{h} = \frac{(a + 2h - a)}{h(\sqrt{a + 2h} + \sqrt{a})} = \frac{2h}{h(\sqrt{a + 2h} + \sqrt{a})}$
## Step2: Cancel h and substitute h=0
$\lim_{h 	o 0} \frac{2}{\sqrt{a + 2h} + \sqrt{a}} = \frac{2}{2\sqrt{a}} = \frac{1}{\sqrt{a}}$
# Answer for 54:
$\frac{1}{\sqrt{a}}$

# Explanation for 55:
## Step1: Expand numerator
$(x + a)^3 - a^3 = x^3 + 3x^2a + 3xa^2 + a^3 - a^3 = x^3 + 3x^2a + 3xa^2$
## Step2: Divide by x and substitute x=0
$\lim_{x 	o 0} (x^2 + 3xa + 3a^2) = 3a^2$
# Answer for 55:
$3a^2$

# Explanation for 56:
## Step1: Simplify numerator
$\frac{1}{h} - \frac{1}{a} = \frac{a - h}{ah}$
## Step2: Combine with denominator
$\frac{\frac{a - h}{ah}}{h - a} = \frac{-(h - a)}{ah(h - a)} = -\frac{1}{ah}$
## Step3: Substitute h=a
$\lim_{h 	o a} -\frac{1}{ah} = -\frac{1}{a^2}$
# Answer for 56:
$-\frac{1}{a^2}$

# Explanation for 57:
## Step1: Let x=⁴√(1+h), h=x⁴ -1
$\lim_{h 	o 0} \frac{\sqrt[4]{1+h} - 1}{h} = \lim_{x 	o 1} \frac{x - 1}{x^4 - 1}$
## Step2: Factor denominator and cancel (x-1)
$x^4 - 1 = (x - 1)(x^3 + x^2 + x + 1)$, limit = $\frac{1}{1 + 1 + 1 + 1} = \frac{1}{4}$
# Answer for 57:
$\frac{1}{4}$

# Explanation for 58:
## Step1: Let x=⁶√(1+h), h=x⁶ -1
$\lim_{h 	o 0} \frac{\sqrt[3]{1+h} - 1}{\sqrt{1+h} - 1} = \lim_{x 	o 1} \frac{x^2 - 1}{x^3 - 1}$
## Step2: Factor numerator and denominator, cancel (x-1)
$x^2 - 1 = (x - 1)(x + 1)$, $x^3 - 1 = (x - 1)(x^2 + x + 1)$, limit = $\frac{x + 1}{x^2 + x + 1} = \frac{2}{3}$
# Answer for 58:
$\frac{2}{3}$

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QUESTION IMAGE

Question

Explanation:

Step1: Direct substitution x=0

Step1: Substitute h=-2

Step2: Simplify expression

Step1: Substitute t=-1

Step2: Simplify terms

Step1: Expand numerator

Step2: Factor numerator

Step3: Cancel (x - a) and substitute x=a

Answer: