Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find lim x tends to 1 f x , where f x equals 2x square minus 1 for x lesser than 1

24. Find \lim_{x \rightarrow 1} f (x ) , \: \:where \: \: f (x) = \left\{\begin{matrix} x^2 -1 & x \neq 0 \\ -x^2 -1 & x > 1 \end{matrix}\right.

Answers (1)

best_answer

\lim_{x \rightarrow 1} f (x ) , \: \:where \: \: f (x) = \left\{\begin{matrix} x^2 -1 & x \neq 0 \\ -x^2 -1 & x > 1 \end{matrix}\right.

Limit at x=1^+

\lim_{x \rightarrow 1^+} f (x ) = \lim_{x \rightarrow 1} (-x^2-1)=-(1)^2-1=-2

Limit at x=1^-

\lim_{x \rightarrow 1^-} f (x ) = \lim_{x \rightarrow 1} (x^2-1)=(1)^2-1=0

As we can see that Limit at x=1^+ is not equal to Limit at x=1^-,The limit of this function at x = 1 does not exists.

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2027 LIVE: Kerala SSLC dates out; Assam, Bihar, Gujarat Class 10, 12 schedules awaited
anam-khan

CBSE Date Sheet 2027 (OUT) LIVE: Class 10, 12 exam dates out for over 46 lakh students; passing marks
anam-khan

CBSE Class 12 exams clash with Holi on March 22; will board revise date sheet?

Latest Question