Question

If $\displaystyle \lim_{x \to 0} \dfrac{|x|}{\sqrt{x^4 + 4x^2 + 5}} = k$, $\displaystyle \lim_{x \to 0} x^4 \sin\left(\dfrac{1}{3\sqrt{x}}\right) = l$, Then $k + l = $

• A. $0$
• B. $1$
• C. $-1$
• D. $5$

Hint:

When bounded function multiplies a term tending to $0$, the product tends to $0$.

Answer 1

The Correct Option is A

First, evaluate $k = \displaystyle \lim_{x \to 0} \dfrac{|x|}{\sqrt{x^4 + 4x^2 + 5}}$
As $x \to 0$, the denominator tends to $\sqrt{5}$ and numerator tends to $0$
So, $k = \dfrac{0}{\sqrt{5}} = 0$
Now, $l = \displaystyle \lim_{x \to 0} x^4 \sin\left(\dfrac{1}{3\sqrt{x}}\right)$
As $x \to 0$, $\dfrac{1}{\sqrt{x}} \to \infty$ so $\sin(\cdot)$ oscillates in $[-1,1]$
But $x^4 \to 0$, hence $x^4 \cdot \sin(...) \to 0$
So, $l = 0$
Thus, $k + l = 0 + 0 = 0$