<s>[INST] Evaluate $\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}$ Evaluate $$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}$$
My attempt: $$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}=\lim_{x\to+\infty} \frac{3x^2+\cos{\sqrt{x}}}{x^3\sin{\frac{1}{x}}+\frac{1}{x^3}}$$ Let $t=\frac{1}{x}$, then
$$\lim_{x\to+\infty} \frac{3x^2+\cos{\sqrt{x}}}{x^3\sin{\frac{1}{x}}+\frac{1}{x^3}}=\lim_{t\to0^+} \frac{\frac{3}{t^2}+\cos{\frac{1}{\sqrt{t}}}}{\frac{1}{t^3}\sin{t}+t^3}$$
That's where I got stuck. I think this substitution didn't help much. Maybe there's a way to apply the squeeze theorem, but it's not so obvious. Hint, please?
 [/INST] $$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}=\lim_{x\to+\infty} \frac{3x^3(1+\frac{x\cos{\sqrt{x}}}{3x^3}) }{x^3 \left( \frac{\sin{\frac{1}{x}}}{\frac{1}{x}}  +\frac{1}{x^3} \right)} = 3$$
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