Calculus Examples

$\frac{lim_{x \to 0} \sqrt{1 + \tan (x)} - \sqrt{1 \sin (x)}}{x^{3}}$

Step 1

Evaluate the limit.

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Step 1.1

Multiply $\sin (x)$ by $1$ .

$\frac{lim_{x \to 0} \sqrt{1 + \tan (x)} - \sqrt{\sin (x)}}{x^{3}}$

Step 1.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{lim_{x \to 0} \sqrt{1 + \tan (x)} - lim_{x \to 0} \sqrt{\sin (x)}}{x^{3}}$

Step 1.3

Move the limit under the radical sign.

$\frac{\sqrt{lim_{x \to 0} 1 + \tan (x)} - lim_{x \to 0} \sqrt{\sin (x)}}{x^{3}}$

Step 1.4

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

$\frac{\sqrt{lim_{x \to 0} 1 + lim_{x \to 0} \tan (x)} - lim_{x \to 0} \sqrt{\sin (x)}}{x^{3}}$

Step 1.5

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

$\frac{\sqrt{1 + lim_{x \to 0} \tan (x)} - lim_{x \to 0} \sqrt{\sin (x)}}{x^{3}}$

Step 1.6

Move the limit inside the trig function because tangent is continuous.

$\frac{\sqrt{1 + \tan (lim_{x \to 0} x)} - lim_{x \to 0} \sqrt{\sin (x)}}{x^{3}}$

Step 2

Since $\sin (x)$ is negative near $0$ , and the radical is not defined for negative values, the limit does not exist.

$\frac{\sqrt{1 + \tan (lim_{x \to 0} x)} - Does not exist}{x^{3}}$

Step 3

The limit does not exist.

$Does not exist$