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.