$$ \lim _ { \Delta x ightarrow 0 } \frac { \sqrt { an ( x + \Delta x ) } - \sqrt { an x } } { \Delta x } $$

Question

$$ \lim _ { \Delta x ightarrow 0 } \frac { \sqrt { 	an ( x + \Delta x ) } - \sqrt { 	an x } } { \Delta x } $$

UpStudy AI · Accepted Answer

To evaluate the limit

$$
\lim _ { \Delta x \rightarrow 0 } \frac { \sqrt { \tan ( x + \Delta x ) } - \sqrt { \tan x } } { \Delta x },
$$

we can recognize that this expression resembles the definition of the derivative of a function. Specifically, we can define the function

$$
f(x) = \sqrt{\tan x}.
$$

Then, the limit can be rewritten as

$$
\lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = f'(x).
$$

To find $$ f'(x) $$, we will use the chain rule. First, we need to find the derivative of $$ f(x) = \sqrt{\tan x} $$.

Using the chain rule, we have:

$$
f'(x) = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x,
$$

where $$ \sec^2 x $$ is the derivative of $$ \tan x $$.

Thus, we can express $$ f'(x) $$ as:

$$
f'(x) = \frac{\sec^2 x}{2\sqrt{\tan x}}.
$$

Now, substituting this back into our limit, we find:

$$
\lim _ { \Delta x \rightarrow 0 } \frac { \sqrt { \tan ( x + \Delta x ) } - \sqrt { \tan x } } { \Delta x } = \frac{\sec^2 x}{2\sqrt{\tan x}}.
$$

Therefore, the final result is:

$$
\boxed{\frac{\sec^2 x}{2\sqrt{\tan x}}}.
$$
The limit is $$ \frac{\sec^2 x}{2\sqrt{\tan x}} $$.

\( \lim _ { \Delta x \rightarrow 0 } \frac { \sqrt { \tan ( x + \Delta x ) } - \sqrt { \tan x } } { \Delta x } \)