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Question 165
Find the limit limx→0e4x−1x\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 } { x }limx→0xe4x−1
A) limx→0e4x−1x=2\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 } { x } = 2limx→0xe4x−1=2 B) limx→0e4x−1x=−4\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 } { x } = - 4limx→0xe4x−1=−4 C) limx→0e4x−1x=4\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 } { x } = 4limx→0xe4x−1=4 D) limx→0e4x−1x=−6\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 } { x } = - 6limx→0xe4x−1=−6 E) limx→0e4x−1x=6\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 } { x } = 6limx→0xe4x−1=6
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