Question:
If the function \( f(x) \) defined by \( f(x)=\begin{cases}\dfrac{\log_{e}(1+5x)-\log_{e}(1-2x)}{x}, & x \neq 0 \\ l, & x=0\end{cases} \) is continuous at \( x=0 \), then the value of \( l \) is :
If the function \( f(x) \) defined by \( f(x)=\begin{cases}\dfrac{\log_{e}(1+5x)-\log_{e}(1-2x)}{x}, & x \neq 0 \\ l, & x=0\end{cases} \) is continuous at \( x=0 \), then the value of \( l \) is :
Show Hint
Continuity at 0 needs \( l=\lim_{x\to0}f(x) \); use \( \lim_{x\to0}\ln(1+kx)/x=k \).
Updated On: Nov 12, 2026
- 3
- 2
- 5
- 7
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The Correct Option is D
Solution and Explanation
For continuity at \( x=0 \), \( l=\lim_{x\to0}\dfrac{\ln(1+5x)-\ln(1-2x)}{x} \).
Split: \( \lim_{x\to0}\dfrac{\ln(1+5x)}{x}=5 \) and \( \lim_{x\to0}\dfrac{\ln(1-2x)}{x}=-2 \).
So \( l=5-(-2)=7 \).
Answer: (4) 7
Split: \( \lim_{x\to0}\dfrac{\ln(1+5x)}{x}=5 \) and \( \lim_{x\to0}\dfrac{\ln(1-2x)}{x}=-2 \).
So \( l=5-(-2)=7 \).
Answer: (4) 7
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