The correct option is D does not exist L.H.L.=lim_x→(π2)^ tan x=+∞ R.H.L.=lim_x→(π2)^+tan x= ∞ Clearly left hand limit ≠ right hand limit.Hence ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Existence of Limit

limx→π/2tan x...

Question

$lim x \to \frac{π}{2} tan x =$

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{π}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

does not exist

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D does not exist
$L.H.L. = lim x \to {(\frac{π}{2})}^{-} tan x = + \infty$
$R.H.L. = lim x \to {(\frac{π}{2})}^{+} tan x = - \infty$
Clearly left hand limit $\neq$ right hand limit.
Hence given limit does not exist.

Suggest Corrections

Similar questions

lim x \to 0 \frac{\sqrt{1 - cos 2 x}}{x}

| sin x | = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} sin x; & 0 < x < \frac{π}{2} - sin x; & - \frac{π}{2} < x < 0 \end{matrix}

{lim}_{x \to π / 4} \frac{{sec}^{2} x - 2 tan x}{1 + cos 4 x}

{lim}_{x \to \frac{π}{2}} tan x

\lim_{x \to π / 4} \frac{\sec^{2} x - 2}{\tan x - 1} is

\sqrt{2}

l i m x \to \frac{π}{2} tan x

Join BYJU'S Learning Program