Let f ( x ) = 4 x − π 1 − t a n x , x  = 4 π , x ∈ [ 0 , 2 π ] . If f ( x ) is continuous in [ 0 , 2 π ] , then f ( 4 π ) is

f ( x ) = 4 x − π 1 − t a n x ⇒ lim x → 4 π 4 x − π 1 − t a n x = − 2 1

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Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsTrigonometric and Inverse Trigonometric limitsEasy2 minPYQ_2004

MathematicsEasysingle choice

Let $f (x) = \frac{1 - t a n x}{4 x - π}, x \neq = \frac{π}{4}, x \in [0, \frac{π}{2}]$ . If $f (x)$ is continuous in $[0, \frac{π}{2}]$ , then $f (\frac{π}{4})$ is

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Solution:

$f (x) = \frac{1 - t a n x}{4 x - π} \Rightarrow lim_{x \to \frac{π}{4}} \frac{1 - t a n x}{4 x - π} = - \frac{1}{2}$

Stream:JEESubject:MathematicsTopic:LimitsSubtopic:Trigonometric and Inverse Trigonometric limits

⏱ 2mℹ️ Source: PYQ_2004

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