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Let $\displaystyle l=\lim_{x\rightarrow 0}\int_{0}^{x}\frac{(1+\cos t)^2}{x}dt$ and $\displaystyle m=\lim_{x\rightarrow \infty}\int_{0}^{x}\frac{(1+\cos t)^2}{x}dt\;,$ Then $l+m=$

$\bf{My\; Try::}$ Given $\displaystyle l=\lim_{x\rightarrow 0}\int_{0}^{x}\frac{(1+\cos t)^2}{x}dt\; \left(\frac{0}{0}\right)$ Form (Using L Hopital Rule)

We Get $$l=\lim_{x\rightarrow 0}\frac{(1+\cos x)^2}{1} = 4$$

Now How can I solve second Limit, Help required, Thanks

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    $\begingroup$ Just integrate? $\int_0^x (1+\cos t)^2\,dt=3x/2+2\sin x+\sin(2x)/4$. $\endgroup$ – mickep May 27 '16 at 6:40
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$$\lim_{x\rightarrow\infty}\frac{1}{x}\int_{0}^{x}(1+\cos^2t+2\cos{t}) dt$$ now write $\cos^2t=\frac{\cos2t+1}{2}$ and solve the integral. note though $\cos(\infty)$ is variable its bounded by $(-1,1)$

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