Let
f ( x ) = b 0 x n + b 1 x n + 1 + ⋯ + b l x n + l a 0 x m + a 1 x m + 1 + ⋯ + a k x m + k ,
where a 0 = 0 , b 0 = 0 , and m , n ∈ N .
Then given (A), (B), (C), or (D), x → 0 lim f ( x ) equals which of (1), (2), (3), and (4)?
Match the columns:
Column-I | Column-II |
(A) if m > n | (1) ∞ |
(B) if m = n | (2) − ∞ |
(C) if m < n , n − m is even, and b 0 a 0 > 0 | (3) b 0 a 0 |
(D) if m < n , n − m is even, and b 0 a 0 < 0 | (4) 0 |
Note: For example, if (A) correctly matches (1), (B) with (2), (C) with (3), and (D) with (4), then answer as 1234.
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I think we say limit of function exists at a point when both left hand limit and right hand limit of that function are finite and are equal. So for case 3 and 4 does the limit exist at x =0?
1st case simplifies to lim(x->0)ax/b = 0
2nd case simplifies to lim(x->0) (a 0*x^m-1)/(b 0*x^m-1) = a 0/b 0
3rd case simplifies to lim(x->0) a 0/b 0*x^2 = +infinity
4th case takes the remaining option
Can u explain elaborately
@Terrell Bombb please explain how you simplified
you really helped me sooo fucking much dude well done.
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