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The value of lim(xrarrpi//4) (tan^(3)x-t...

The value of `lim_(xrarrpi//4) (tan^(3)x-tanx)/(cos(x+(pi)/(4)))` is

A

8

B

4

C

`-8`

D

`-2`

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The correct Answer is:
To solve the limit \( \lim_{x \to \frac{\pi}{4}} \frac{\tan^3 x - \tan x}{\cos\left(x + \frac{\pi}{4}\right)} \), we will follow these steps: ### Step 1: Substitute \( x = \frac{\pi}{4} \) First, we substitute \( x = \frac{\pi}{4} \) into the expression to check if it results in an indeterminate form. \[ \tan\left(\frac{\pi}{4}\right) = 1 \] So, the numerator becomes: \[ \tan^3\left(\frac{\pi}{4}\right) - \tan\left(\frac{\pi}{4}\right) = 1^3 - 1 = 0 \] Now, for the denominator: \[ \cos\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \] Since both the numerator and denominator are 0, we have the indeterminate form \( \frac{0}{0} \). ### Step 2: Apply L'Hôpital's Rule Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately. #### Derivative of the Numerator The numerator is \( \tan^3 x - \tan x \). We will differentiate it: \[ \frac{d}{dx}(\tan^3 x) = 3\tan^2 x \sec^2 x \] \[ \frac{d}{dx}(\tan x) = \sec^2 x \] Thus, the derivative of the numerator is: \[ 3\tan^2 x \sec^2 x - \sec^2 x = \sec^2 x (3\tan^2 x - 1) \] #### Derivative of the Denominator The denominator is \( \cos\left(x + \frac{\pi}{4}\right) \). The derivative is: \[ \frac{d}{dx}\left(\cos\left(x + \frac{\pi}{4}\right)\right) = -\sin\left(x + \frac{\pi}{4}\right) \] ### Step 3: Rewrite the Limit Now we can rewrite the limit using the derivatives: \[ \lim_{x \to \frac{\pi}{4}} \frac{3\tan^2 x \sec^2 x - \sec^2 x}{-\sin\left(x + \frac{\pi}{4}\right)} \] ### Step 4: Substitute \( x = \frac{\pi}{4} \) Again Now we substitute \( x = \frac{\pi}{4} \) again: 1. Calculate \( \tan\left(\frac{\pi}{4}\right) = 1 \) and \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \). So, the numerator becomes: \[ 3(1^2)(\sqrt{2})^2 - (\sqrt{2})^2 = 3 \cdot 1 \cdot 2 - 2 = 6 - 2 = 4 \] 2. For the denominator: \[ -\sin\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{2}\right) = -1 \] ### Step 5: Final Calculation Now we can compute the limit: \[ \lim_{x \to \frac{\pi}{4}} \frac{4}{-1} = -4 \] ### Conclusion Thus, the value of the limit is: \[ \boxed{-4} \]

To solve the limit \( \lim_{x \to \frac{\pi}{4}} \frac{\tan^3 x - \tan x}{\cos\left(x + \frac{\pi}{4}\right)} \), we will follow these steps: ### Step 1: Substitute \( x = \frac{\pi}{4} \) First, we substitute \( x = \frac{\pi}{4} \) into the expression to check if it results in an indeterminate form. \[ \tan\left(\frac{\pi}{4}\right) = 1 ...
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