Math / Calculus

Question29. $\lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+9}-3}{x^{2}}$
30. $\lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+4}-2}{x}$
31. $\lim _{x \rightarrow 0} \frac{1-\sqrt{1-x^{2}}}{x^{2}}$
32. $\lim _{x \rightarrow 0} \frac{\sqrt{2-x}-\sqrt{2}}{2 x}$

Studdy Solution

STEP 1

1. The expression involves a square root, which suggests rationalization might be useful.
2. Direct substitution of $x = 0$ results in an indeterminate form.

STEP 2

1. Rationalize the numerator.
2. Simplify the expression.
3. Evaluate the limit.
4. 1. Rationalize the numerator.
2. Simplify the expression.
3. Evaluate the limit.
STEP_1: High_Level_Step: 1 Rationalize the numerator by multiplying the numerator and the denominator by the conjugate:
$\frac{\sqrt{x^{2}+4}-2}{x} \cdot \frac{\sqrt{x^{2}+4}+2}{\sqrt{x^{2}+4}+2}$
This results in:
$\frac{(x^{2}+4) - 4}{x(\sqrt{x^{2}+4}+2)} = \frac{x^{2}}{x(\sqrt{x^{2}+4}+2)}$
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2 Cancel $x$ in the numerator and denominator:
$\frac{x}{\sqrt{x^{2}+4}+2}$
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3 Evaluate the limit as $x \rightarrow 0$ :
$\lim_{x \rightarrow 0} \frac{x}{\sqrt{x^{2}+4}+2} = \frac{0}{2+2} = 0$
High_Level_Step_Completed: TRUE
The value of the limit is:
$\boxed{0}$
### Problem 31: $\lim _{x \rightarrow 0} \frac{1-\sqrt{1-x^{2}}}{x^{2}}$
_ASSUMPTIONS_:
1. The expression involves a square root, suggesting rationalization.
2. Direct substitution of $x = 0$ results in an indeterminate form.
_HIGH_LEVEL_APPROACH_:
1. Rationalize the numerator.
2. Simplify the expression.
3. Evaluate the limit.
STEP_1: High_Level_Step: 1 Rationalize the numerator by multiplying the numerator and the denominator by the conjugate:
$\frac{1-\sqrt{1-x^{2}}}{x^{2}} \cdot \frac{1+\sqrt{1-x^{2}}}{1+\sqrt{1-x^{2}}}$
This results in:
$\frac{1 - (1-x^{2})}{x^{2}(1+\sqrt{1-x^{2}})} = \frac{x^{2}}{x^{2}(1+\sqrt{1-x^{2}})}$
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2 Cancel $x^{2}$ in the numerator and denominator:
$\frac{1}{1+\sqrt{1-x^{2}}}$
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3 Evaluate the limit as $x \rightarrow 0$ :
$\lim_{x \rightarrow 0} \frac{1}{1+\sqrt{1-x^{2}}} = \frac{1}{1+1} = \frac{1}{2}$
High_Level_Step_Completed: TRUE
The value of the limit is:
$\boxed{\frac{1}{2}}$
### Problem 32: $\lim _{x \rightarrow 0} \frac{\sqrt{2-x}-\sqrt{2}}{2x}$
_ASSUMPTIONS_:
1. The expression involves a square root, suggesting rationalization.
2. Direct substitution of $x = 0$ results in an indeterminate form.
_HIGH_LEVEL_APPROACH_:
1. Rationalize the numerator.
2. Simplify the expression.
3. Evaluate the limit.
STEP_1: High_Level_Step: 1 Rationalize the numerator by multiplying the numerator and the denominator by the conjugate:
$\frac{\sqrt{2-x}-\sqrt{2}}{2x} \cdot \frac{\sqrt{2-x}+\sqrt{2}}{\sqrt{2-x}+\sqrt{2}}$
This results in:
$\frac{(2-x) - 2}{2x(\sqrt{2-x}+\sqrt{2})} = \frac{-x}{2x(\sqrt{2-x}+\sqrt{2})}$
High_Level_Step_Completed: TRUE
STEP_2: High_Level_Step: 2 Cancel $x$ in the numerator and denominator:
$\frac{-1}{2(\sqrt{2-x}+\sqrt{2})}$
High_Level_Step_Completed: TRUE
STEP_3: High_Level_Step: 3 Evaluate the limit as $x \rightarrow 0$ :
$\lim_{x \rightarrow 0} \frac{-1}{2(\sqrt{2-x}+\sqrt{2})} = \frac{-1}{2(2)} = \frac{-1}{4}$
High_Level_Step_Completed: TRUE
The value of the limit is:
$\boxed{-\frac{1}{4}}$

STEP 3

Rationalize the numerator by multiplying the numerator and the denominator by the conjugate:
$\frac{\sqrt{x^{2}+9}-3}{x^{2}} \cdot \frac{\sqrt{x^{2}+9}+3}{\sqrt{x^{2}+9}+3}$
This results in:
$\frac{(x^{2}+9) - 9}{x^{2}(\sqrt{x^{2}+9}+3)} = \frac{x^{2}}{x^{2}(\sqrt{x^{2}+9}+3)}$

STEP 4

Cancel $x^{2}$ in the numerator and denominator:
$\frac{1}{\sqrt{x^{2}+9}+3}$

STEP 5

Evaluate the limit as $x \rightarrow 0$ :
$\lim_{x \rightarrow 0} \frac{1}{\sqrt{x^{2}+9}+3} = \frac{1}{3+3} = \frac{1}{6}$
The value of the limit is:
$\boxed{\frac{1}{6}}$
### Problem 30: $\lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+4}-2}{x}$
_ASSUMPTIONS_:
1. The expression involves a square root, suggesting rationalization.
2. Direct substitution of $x = 0$ results in an indeterminate form.

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