Existence of Limit
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Let f(x)=x(−1)[1x].x≠0, where [x] denotes the greatest integer less than or equal to x. then limx→0f(x)
Does not exist
is equal to 2
is equal to 0
is equal to -1
- (0, π)
- [−π2, π2]
- (−π2, π2)
- (−π2, π2)−{0}
(where [.] is greatest integer function)
- Right hand limit (RHL) is 1.
- Left hand limits (LHL) is −1.
- Both LHL and RHL is 0.
- Both LHL and RHL is 1.
Let f(x)={x2k(x2−4)2−xwhen x is an int eger otherwise then limx→2f(x)
exists only when k=1
exists for every real k
exists for every real k = 1
does not exist
limx→3([x−3]+[3−x]−x), where[.]denote the greatest integer function, is equal to:
4
-4
0
Does not exist
The value of limx→π2[sin−1sinx], [x] is the greatest integer function of x, is
1
π2
0
12
limx→1f(x)=
limx→3([x−3]+[3−x]−x), where[.]denote the greatest integer function, is equal to:
4
-4
0
Does not exist
limx→0 √1−cos2x√2x is
(JEE 2002)
1
-1
Zero
Does not exist
- −1
- 0
- 1
- does not exist
limx→0√1−cos2x√2xis
λ
-1
zero
Does not exist
- −1
- 0
- 1
- does not exist
consider the function
f(x)=⎧⎪⎨⎪⎩a+bx, x<14, x=1b−ax, x>1
If limx→1 f(x)=f(1), then the values of a and b are
a = 3, b = 1
a = 1, b = 3
a = 0, b = 4
a = 4, b = 0
- x2−6x+9=0
- x2−7x+8=0
- x2−14x+49=0
- x2−10x+21=0
- 2
- 0
- 3
- 1