Question
Download Solution PDFWhat is \(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\) equal to ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) + \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)
\(\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac { (a^x - 1) }{x} = \log a\)
log mn = n log m
Calculation:
\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1+ 3^{-x}-1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} -1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^{-1})^x -1}{x}\\= \log 3 + \log (3^{-1})\\= \log 3 - \log 3\\=0\)
Last updated on May 30, 2025
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