Question:
Evaluate the Given limit: \(\lim_{x\rightarrow 0}\) \(\frac{ax+xcos\,x}{b\,sin\,x}\)
Evaluate the Given limit: \(\lim_{x\rightarrow 0}\) \(\frac{ax+xcos\,x}{b\,sin\,x}\)
Updated On: Oct 23, 2023
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Solution and Explanation
\(\lim_{x\rightarrow 0} \frac{ax+xcos\,x}{b\,sin\,x}\)
At x = 0, the value of the given function takes the form 0/0.
\(\lim_{x\rightarrow 0}\) \(\frac{cos^2x-1}{cos\,x-1}\) =\(\frac{1}{b}\)\(\lim_{x\rightarrow 0}\) \(\frac{x(a+cos\,x)}{sin\,x}\)
=\(\frac{1}{b}\)\(\times\)\(\lim_{x\rightarrow 0}\)(\(\frac{x}{sin\,x}\)) \(\times\)\(\lim_{x\rightarrow 0}\) (a + cosx)
=\(\frac{1}{b}\) \(\times\) (\(\lim_{x\rightarrow 0}\)\(\frac{1}{\frac{sin\,x}{x}}\)) \(\times\) \(\lim_{x\rightarrow 0}\) (a + cosx)
= \(\frac{1}{b}\) \(\times\) (a + cos0) [\(\lim_{x\rightarrow 0}\) \(\frac{sin\,x}{x}\) = 1]
= \(a+\frac{1}{b}\)
At x = 0, the value of the given function takes the form 0/0.
\(\lim_{x\rightarrow 0}\) \(\frac{cos^2x-1}{cos\,x-1}\) =\(\frac{1}{b}\)\(\lim_{x\rightarrow 0}\) \(\frac{x(a+cos\,x)}{sin\,x}\)
=\(\frac{1}{b}\)\(\times\)\(\lim_{x\rightarrow 0}\)(\(\frac{x}{sin\,x}\)) \(\times\)\(\lim_{x\rightarrow 0}\) (a + cosx)
=\(\frac{1}{b}\) \(\times\) (\(\lim_{x\rightarrow 0}\)\(\frac{1}{\frac{sin\,x}{x}}\)) \(\times\) \(\lim_{x\rightarrow 0}\) (a + cosx)
= \(\frac{1}{b}\) \(\times\) (a + cos0) [\(\lim_{x\rightarrow 0}\) \(\frac{sin\,x}{x}\) = 1]
= \(a+\frac{1}{b}\)
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