Question
Download Solution PDFWhat is \(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\) equal to?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)
\(\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} = 1\\\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+x) }{x}\)
Calculation:
We have to find the value of \(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\)
As we know, \(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)
\(\therefore \rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}= \rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} × \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1-x) }{x}\)
= 1 × \(\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{x}\)
= \(\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{-(-x)}\)
= \(-1 × \rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{(-x)}\)
= -1 × 1
= -1
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