Question
Solution
We want to evaluate the limit: 
\ As x→0, the numerator approaches e0 − 1 − 0 = 1 – 1 – 0 = 0, and the denominator approaches 02 = 0. So, we have an indeterminate form of type 0/0and we can use L'Hôpital's Rule. Applying L'Hôpital's Rule once: 
As x→0, the numerator approaches e0 −1 = 1 – 1 = 0, and the denominator approaches 2(0)=0. We still have an indeterminate form of type 0/0​, so we apply L'Hôpital's Rule again. Applying L'Hôpital's Rule a second time: 
In each of the questions below are given some statements followed by two conclusions. You have to take the given statements to be true even if they see...
Statement:  J > K ≥ C = D ≤ Y < Z
Conclusions:
I. Z > C
II. J ≥ Y
III. J > D
IV. C > Z
...In the following question the relationship between different elements is given in the statements followed by three conclusions I, II and III. Read the ...
Statement: H > G = M > S; GÂ `>=` Â T > L; MÂ `<=` Â F <Â Â U
Conclusion: Â I. F > SÂ Â Â Â Â Â Â Â Â Â Â II. T < H
...Statements:Â Â Â Â Â Â T @ V % Z #Â C & B $ SÂ # E; W $ Z @ C
Conclusions :Â Â Â Â Â I. E @ ZÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â II. S # WÂ Â Â ...
Statements: B ≥ C > D; B < E > J; G > A ≥ H > J
Conclusion:
I. D ≤ A
II. G > C
Statements: Q © E, S % C, E $ S, C @ AÂ
Conclusions:Â
I. A © CÂ
II. S % AÂ
III. C © Q
Statement: L ≥ M ≤ R = S; M > N ≥ P
Conclusions: I. P ≤ M II. L > N
Statements:
A ≥ T > I, D > I, U = P ≥ I
Conclusions:
I. U > D
II. A > I
Statements: B @ E, E $ Y, Y & I, I % C
Conclusions: I. E @ IÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â II. C & B
...
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