The limit of sin(√x)/x as x approaches 0 DOES NOT exist, that is, the lim_{x→0} sin(\sqrt{x})/x formula is given by

$\lim\limits_{x \to 0} \dfrac{\sin \sqrt{x}}{x}=$ not exist.

Let us learn how to find the limit of sin root x divided by x when x tends to 0.

## Find Limit of sin(√x)/x when x→0

We have

$\lim\limits_{x \to 0} \dfrac{\sin \sqrt{x}}{x}$

= $\lim\limits_{x \to 0} \dfrac{\sin \sqrt{x}}{\sqrt{x}} \times \dfrac{1}{\sqrt{x}}$

= $\lim\limits_{x \to 0} \dfrac{\sin \sqrt{x}}{\sqrt{x}} \times \lim\limits_{x \to 0} \dfrac{1}{\sqrt{x}}$ by the product rule of limits.

[Let √x=z, so that z→0 as x→0]

= $\lim\limits_{z \to 0} \dfrac{\sin z}{z} \times \lim\limits_{x \to 0} \dfrac{1}{\sqrt{x}}$

= 1 × 1/0 using the formula lim_{x→0} sinx/x =1.

= does not exist.

So the limit of sin(√x)/x does not exist as x→0.

**More reading:**

Lim_{x→0} x/sinx | Lim_{x→∞} sinx/x |

Lim_{x→0} sin(1/x) | Lim_{x→0} sin3x/sin2x |

Lim_{x→0} tanx/x | Lim_{x→0} (cosx-1)/x |

## Limit of sin(√x)/2x as x→0

The given limit can be written as follows:

$\lim\limits_{x \to 0} \dfrac{\sin \sqrt{x}}{2x}$ = $\dfrac{1}{2} \lim\limits_{x \to 0} \dfrac{\sin \sqrt{x}}{x}$

So by the above limit we prove, we conclude that the limit of sin(√x)/2x as x→0 does not exist.

**More Limits:**

## FAQs

### Q1: What is the limit of sin(√x)/x when x→0?

Answer: The limit of sin(√x)/x when x→0 does not exist.