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Grade 12Algebra

Solve the question given in the attached image............................................................
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Profile image of Sarvesh
7 Years agoGrade 12
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1 Answer

Profile image of Omkar Doiphode
7 Years ago
\lim_{x\rightarrow \infty}(1+xe^{\frac{-1}{x^{2}}}sin(\frac{1}{x^{4}}))
= 1 + \lim_{x\rightarrow \infty}(xe^{\frac{-1}{x^{2}}}sin(\frac{1}{x^{4}}))
= 1+ \lim_{x\rightarrow \infty}(xe^{\frac{-1}{x^{2}}}) \lim_{x\rightarrow \infty}(sin(\frac{1}{x^{4}})
=1+0.\lim_{x\rightarrow \infty}(sin(\frac{1}{x^{4}})
 
Note that sine function is a bounded function, i.e. its value will always lie in the interval [-1,1]
\lim_{x\rightarrow \infty}(sin(\frac{1}{x^{4}}) \in [-1,1]
Hence value of given limit is 1+0=1