Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
shefali sharma Grade: 12

check continuity and differentiability

f(x)= {1/1+ e^1/x} , x is not equal to 0

0   , x=0

7 years ago

## Answers : (1)

Jitender Singh
IIT Delhi
158 Points

Ans:
Checking the continuity:
Left hand limit(LHL) = Right hand limit(RHL) = f(0)
$LHL = \lim_{h\rightarrow 0}\frac{1}{1+e^{\frac{-1}{h}}}$
$LHL = \lim_{h\rightarrow 0}\frac{e^{\frac{1}{h}}}{1+e^{\frac{1}{h}}}$
$LHL = \lim_{h\rightarrow 0}\frac{e^{\frac{1}{h}}+1-1}{1+e^{\frac{1}{h}}}$
$LHL = \lim_{h\rightarrow 0}1-\frac{1}{1+e^{\frac{1}{h}}} = 1$
$RHL = \lim_{h\rightarrow 0}\frac{1}{1+e^{\frac{1}{h}}} = 0$
$f(0) = 0$
So, f(x) is not continuous at x = 0.
Similarly, you can check the left hand & right hand derivative.
$\lim_{h\rightarrow 0}\frac{f(-h)}{-h} = \lim_{h\rightarrow 0}\frac{f(h)}{h}$
Thanks & Regards
Jitender Singh
IIT Delhi
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Differential Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 2,900 off
USE CODE: CHENNA
Get extra Rs. 1,696 off
USE CODE: CHENNA