Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Examine the following functions for continuity. f x=x ^ 2 - 25 / x + 5

5 c) Examine the following functions for continuity.

f (x) = \frac{x ^2-25}{x+5}, x \neq -5

Answers (1)

best_answer

Given function is
f(x ) = \frac{x^2-25}{x+5}
For every real number  k , k \neq -5
We gwt,
f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = f(k)
Hence, function  f(x ) = \frac{x^2-25}{x+5} continuous for every real value of x , x \neq -5

Posted by

Gautam harsolia

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

CBSE practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question