Question Find limn→∞n+sin(n2)n+cos(n)
My Approach limn→∞n+sin(n2)n+cos(n)=limn→∞[nn+cos(n)+sin(n2)n+cosn]=limn→∞[11+cos(n)n+sin(n2)n+cosn]
Applying L ' Hospital is not working here
Answer
limn→∞n+sin2(n)n+cos(n)−1=limn→∞sin2(n)−cos(n)n+cos(n)=0,since the numerator is bounded and the denominator tends to +∞. Thereforelimn→∞n+sin2(n)n+cos(n)=1.
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