Question

question 12 evaluate the limit: \\(\lim_{x \to 0} \frac{\sqrt{2x + 81} - 9}{x}\\) question help: submit question

Explanation:

Step1: Rationalize the numerator

Multiply numerator and denominator by the conjugate $\sqrt{2x+81}+9$:
$$\lim_{x \to 0} \frac{\sqrt{2x+81}-9}{x} \cdot \frac{\sqrt{2x+81}+9}{\sqrt{2x+81}+9}$$

Step2: Simplify the numerator

Use $(a-b)(a+b)=a^2-b^2$ to expand numerator:
$$\lim_{x \to 0} \frac{(2x+81)-81}{x(\sqrt{2x+81}+9)}$$

Step3: Cancel common terms

Simplify numerator and cancel $x$:
$$\lim_{x \to 0} \frac{2x}{x(\sqrt{2x+81}+9)} = \lim_{x \to 0} \frac{2}{\sqrt{2x+81}+9}$$

Step4: Substitute $x=0$

Evaluate the limit by plugging in $x=0$:
$$\frac{2}{\sqrt{0+81}+9} = \frac{2}{9+9}$$

Step5: Compute final value

Simplify the denominator:
$$\frac{2}{18} = \frac{1}{9}$$

Answer:

$\frac{1}{9}$