# Evaluating $\lim_{x\to+\infty} \frac{\sqrt{x}\cos{x}+2x^2\sin\left({\frac{1}{x}}\right)}{x-\sqrt{1+x^2}}$

Evaluate $$\lim_{x\to+\infty} \frac{\sqrt{x}\cos{x}+2x^2\sin({\frac{1}{x}})}{x-\sqrt{1+x^2}}$$

My attempt: $$\lim_{x\to+\infty} \frac{\sqrt{x}\cos{x}+2x^2\sin\left({\frac{1}{x}}\right)}{x-\sqrt{1+x^2}}=\lim_{x\to+\infty} \frac{x^2\sqrt{x}\left(\frac{\cos{x}}{x^2}+2\frac{\sin{\frac{1}{x}}}{\sqrt{x}}\right)}{x\left(1-\sqrt{1+\frac{1}{x^2}}\right)}$$$$=\lim_{x\to+\infty} x\sqrt{x}\cdot \frac{\left(\frac{\cos{x}}{x^2}+2\frac{\sin{\frac{1}{x}}}{\sqrt{x}}\right)}{1-\sqrt{1+\frac{1}{x^2}}}$$

Both numerator and denominator tend to zero, while $$x\sqrt{x} \to +\infty$$. Any help is appreciated.

• Multiply and divide by $x+\sqrt{1+x^2}$. Commented Jul 7, 2020 at 2:16

HINT: Let $$x=\frac1t$$ $$\lim_{x\to+\infty} \frac{\sqrt{x}\cos{x}+2x^2\sin({\frac{1}{x}})}{x-\sqrt{1+x^2}}=\lim_{t\to 0}\frac{\sqrt{t}\cos\frac1t+\frac{2}{t}\sin t}{1-\sqrt{t^2+1}}$$ $$=\lim_{t\to 0}\frac{\left(\sqrt{t}\cos\frac1t+2\cdot \frac{\sin t}{t}\right)(1+\sqrt{t^2+1})}{(1-\sqrt{t^2+1})(1+\sqrt{t^2+1})}$$ $$=-\lim_{t\to 0}\frac{\left(\sqrt{t}\cos\frac1t+2\cdot \frac{\sin t}{t}\right)(1+\sqrt{t^2+1})}{t^2}$$