Question

\(\lim_{{y \to 0}} \frac{\sqrt{3} + y^3 - \sqrt{3}}{y^3}\)

• A. \(\frac{1}{2\sqrt{3}}\)
• B. \({2\sqrt{3}}\)
• C. \(\frac{1}{3\sqrt{2}}\)
• D. \({3\sqrt{2}}\)

Answer 1

The Correct Option is A

We have the limit:

\(\lim_{{y \to 0}} \frac{\sqrt{3 + y^3} - \sqrt{3}}{y^3}\)

To evaluate this limit, we can use the conjugate. Multiply the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{3 + y^3} + \sqrt{3}\):

\(\lim_{{y \to 0}} \frac{\sqrt{3 + y^3} - \sqrt{3}}{y^3} \cdot \frac{\sqrt{3 + y^3} + \sqrt{3}}{\sqrt{3 + y^3} + \sqrt{3}}\)

\(\lim_{{y \to 0}} \frac{(3 + y^3) - 3}{y^3(\sqrt{3 + y^3} + \sqrt{3})}\)

\(\lim_{{y \to 0}} \frac{y^3}{y^3(\sqrt{3 + y^3} + \sqrt{3})}\)

Cancel out the \(y^3\) terms:

\(\lim_{{y \to 0}} \frac{1}{\sqrt{3 + y^3} + \sqrt{3}}\)

Now, substitute y = 0:

\(\frac{1}{\sqrt{3 + 0^3} + \sqrt{3}}\)

\(\frac{1}{\sqrt{3} + \sqrt{3}}\)

\(\frac{1}{2\sqrt{3}}\)

Therefore, the limit is \(\frac{1}{2\sqrt{3}}\).

Answer: \(\frac{1}{2\sqrt{3}}\)

Answer 2

We are given the limit:

\[ \lim_{{y \to 0}} \frac{\sqrt{3 + y^3} - \sqrt{3}}{y^3} \]

To evaluate this limit, we will multiply both the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{3 + y^3} + \sqrt{3}\). This gives us:

\[ \lim_{{y \to 0}} \frac{\sqrt{3 + y^3} - \sqrt{3}}{y^3} \cdot \frac{\sqrt{3 + y^3} + \sqrt{3}}{\sqrt{3 + y^3} + \sqrt{3}} \]

Now, simplify the numerator:

\[ \lim_{{y \to 0}} \frac{(3 + y^3) - 3}{y^3(\sqrt{3 + y^3} + \sqrt{3})} \]

This simplifies to:

\[ \lim_{{y \to 0}} \frac{y^3}{y^3(\sqrt{3 + y^3} + \sqrt{3})} \]

Cancel out the \(y^3\) terms:

\[ \lim_{{y \to 0}} \frac{1}{\sqrt{3 + y^3} + \sqrt{3}} \]

Now, substitute \(y = 0\):

\[ \frac{1}{\sqrt{3 + 0^3} + \sqrt{3}} \]

This simplifies to:

\[ \frac{1}{\sqrt{3} + \sqrt{3}} = \frac{1}{2\sqrt{3}} \]

Thus, the limit is:

Answer: \(\frac{1}{2\sqrt{3}}\)