Question

\(\lim\limits_{x\rightarrow0}(\frac{\tan x}{\sqrt{2x+4}-2})\) is equal to

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

The Correct Option is A

We need to find the limit: \(\lim_{x \to 0} \frac{\tan x}{\sqrt{2x+4} - 2}\)

We can rationalize the denominator by multiplying by the conjugate:

\(\lim_{x \to 0} \frac{\tan x}{\sqrt{2x+4} - 2} \cdot \frac{\sqrt{2x+4} + 2}{\sqrt{2x+4} + 2}\)

\(\implies \lim_{x \to 0} \frac{\tan x (\sqrt{2x+4} + 2)}{(2x+4) - 4}\)

\(\implies \lim_{x \to 0} \frac{\tan x (\sqrt{2x+4} + 2)}{2x}\)

We can rewrite this as: \(\lim_{x \to 0} \frac{\tan x}{x} \cdot \frac{\sqrt{2x+4} + 2}{2}\)

We know that \(\lim_{x \to 0} \frac{\tan x}{x} = 1\), so we have:

\(\implies \lim_{x \to 0} 1 \cdot \frac{\sqrt{2x+4} + 2}{2}\)

Now we can substitute x = 0:

\(\implies 1 \cdot \frac{\sqrt{2(0)+4} + 2}{2}\)

\(\implies \frac{\sqrt{4} + 2}{2}\)

\(\implies \frac{2 + 2}{2}\)

\(\implies \frac{4}{2}\)

\(\implies 2\)

Therefore, the correct option is (A) 2.

Answer 2

The given limit is: \[ \lim_{x \to 0} \frac{\tan x}{\sqrt{2x + 4} - 2} \] 

Multiply the numerator and denominator by \( \sqrt{2x + 4} + 2 \): \[ \lim_{x \to 0} \frac{\tan x}{\sqrt{2x + 4} - 2} \cdot \frac{\sqrt{2x + 4} + 2}{\sqrt{2x + 4} + 2} \] \[ \left(\sqrt{2x + 4} - 2\right) \left(\sqrt{2x + 4} + 2\right) = (2x + 4) - 4 = 2x \] \[ \lim_{x \to 0} \frac{\tan x \left( \sqrt{2x + 4} + 2 \right)}{2x} \] \( \tan x \), which is \( \tan x \approx x \) as \( x \to 0 \): \[ \lim_{x \to 0} \frac{x \left( \sqrt{2x + 4} + 2 \right)}{2x} \] The \( x \) terms cancel out: \[ \lim_{x \to 0} \frac{\sqrt{2x + 4} + 2}{2} \] \( x \to 0 \): When \( x = 0 \), we have: \[ \frac{\sqrt{2(0) + 4} + 2}{2} = \frac{\sqrt{4} + 2}{2} = \frac{2 + 2}{2} = \frac{4}{2} = 2 \] 

Thus, the correct option is (A): 2