A) 0
B) \[\frac{1}{2}\]
C) 1
D) \[-\frac{1}{2}\]
Correct Answer: B
Solution :
[b] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+x}-1}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+x}-1}{x}\times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1+x-1}{x\left[ \sqrt{1+x}+1 \right]}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{1+x}+1}=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2}\]
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