Then lim x - 1 fx-A- 2B- 0C- Does not existD- 1

The correct option is D 1 and lim_x→ 1+ f(x) = lim_h→ 0 2 (1+h) = 1 Hence limit of the function is 1 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Standard Limits to Remove Indeterminate Form

then lim x → ...

Question

If f(x)= ${\begin{matrix} x, & when 0\leq x \geq 1 2 - x, & 2-x when 1 \leq x \geq 2 \end{matrix}$
then $l i m_{x \to 1}$ f(x) =

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

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

Does not exist

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

Open in App

Solution

Suggest Corrections

Similar questions

Find the value of $lim x \to 0 \frac{| x |}{x}$

[x]

f (x) = x - [x] - cos x

f^{'} (\frac{π}{2}) =

If f(x) = ${\begin{matrix} x, & when 0\leq x \geq 1 2 - x, & 2-x when 1 \leq x \geq 2 \end{matrix}$
then $l i m_{x \to 1}$ f(x) =

$lim x \to 0 \frac{\sqrt{1 - c o s 2 x}}{\sqrt{2} x}$ is

(JEE 2002)

l i m x \to \infty \frac{s i n (2 + \sqrt{3})^{n} π}{(2 - \sqrt{3})^{n}} (n ϵ N) =

Join BYJU'S Learning Program