Basic maths-calculus-Q8

+1 vote

Value  lim_{x\rightarrow 0} \frac{log_e(1+3x)}{e^{2x}-1}

(A) \frac{3}{2}

(B) 0

(C) \frac{1}{2}

(D) \frac{2}{3}

asked May 19 in Basic Maths by gbeditor (44,500 points)
reshown May 24 by gbeditor

1 Answer

0 votes
 
Best answer

After substituting the values of 0 we are getting 0/0 inderminant form 

So we can apply L-hopital rule

\lim_{x\rightarrow 0}log(1+3x)/e^{2x}-1\\ \\ \lim_{x\rightarrow 0}(3/1+3x)/e^{2x}*2\\ \\ \lim_{x\rightarrow 0}(3/1+0)/1*2=3/2\\ \\ \therefore Answer\ should\ be\ 3/2

answered May 27 by deepak-gatebook (112,390 points)
Answer:
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