# 2nd PUC Basic Maths Question Bank Chapter 17 Limit and Continuity of a Function

Students can Download Basic Maths Question Bank Chapter 17 Limit and continuity of a function Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Basic Maths Question Bank Chapter 17 Limit and Continuity of a Function

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

### 2nd PUC Basic Maths Limits and Continuity of a Function Five or Six Marks Questions and Answers

Question 1.

(i.e,n is +ve, -ve and a fraction.)

Problems from standand limits with solution. (One or Two M. Q.)

Evaluate the soloving:

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Put y = x – 1
y + 1 =x
As x → 1
y → 40

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Question 21.

Question 22.

Continuity of a function:
A function y =fx) is said to be continuous at x = a if f(a) exists and ¡s equal to f(a)
i.e. $$\lim _{x \rightarrow a}$$ f(x) = f(a)

OR

A function y = f(x) is said to be continuous at x = a. If $$\lim _{x \rightarrow a^{+}}$$ f(x) = $$\lim _{x \rightarrow a^{-}}$$ = f(x) = f(a) i.e., R.H.L. = L.H.L = f(a).

### 2nd PUC Basic Maths Continuity of a Function Three or Four Marks Questions and Answers

Question 1.
S.T. the function f(x) = |x| is continuous at x = 0.

Question 2.

Question 3.

Question 4.
Define f(0) So that f(x) = $$\frac{x}{1-\sqrt{1-x}}$$ becomes continous at x = 0 and f(0) = 2.
f(0) = $$\frac{0}{1-\sqrt{1-0}}=\frac{0}{1-1}=\frac{0}{0}$$ form, which is inditerminate hence rationalize the D.N.R.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.