Question

If
\[ \lim_{x\to0}\frac{|x|}{\sqrt{x^4+4x^2+5}}=k, \] and
\[ \lim_{x\to0}x^4\sin\left(\frac{1}{3\sqrt{x}}\right)=l, \] then \(k+l=\)

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(5\)

Hint:

If a bounded trigonometric function is multiplied by a term tending to \(0\), then the whole product tends to \(0\).

Answer 1

The Correct Option is A

Step 1: Evaluate \(k\).
We have
\[ k=\lim_{x\to0}\frac{|x|}{\sqrt{x^4+4x^2+5}} \]
As \(x\to0\),
\[ |x|\to0 \]
and
\[ \sqrt{x^4+4x^2+5}\to\sqrt5 \]
Therefore,
\[ k=\frac{0}{\sqrt5} \]
\[ k=0 \]

Step 2: Evaluate \(l\).
We have
\[ l=\lim_{x\to0}x^4\sin\left(\frac{1}{3\sqrt{x}}\right) \]
Since the sine function is always bounded,
\[ -1\leq \sin\left(\frac{1}{3\sqrt{x}}\right)\leq 1 \]
Multiplying by \(x^4\),
\[ -x^4\leq x^4\sin\left(\frac{1}{3\sqrt{x}}\right)\leq x^4 \]
As \(x\to0\),
\[ -x^4\to0 \] and
\[ x^4\to0 \]
Hence, by squeeze theorem,
\[ l=0 \]

Step 3: Find \(k+l\).
\[ k+l=0+0 \]
\[ =0 \]

Step 4: Final conclusion.
Hence,
\[ \boxed{0} \]