 # I need help calculating this limit: $$\displaystyle\lim_{{{x}\to\pi}}{\frac{{{\cos{{2}}}{x}+{2}{\left(\pi-{x}\right)}{\sin{{x}}}-{1}}}{{{\left({\cos{{2}}}{x}+{\cos{{x}}}\right)}^{{2}}}}}$$ I tried Samara Richard 2022-04-01 Answered
I need help calculating this limit:
$\underset{x\to \pi }{lim}\frac{\mathrm{cos}2x+2\left(\pi -x\right)\mathrm{sin}x-1}{{\left(\mathrm{cos}2x+\mathrm{cos}x\right)}^{2}}$
I tried to use L'Hospital's rule, but didn't get any result. Can you help me please?
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Let $x=t+\pi$
Thus,
$\underset{x\to \pi }{lim}\frac{\mathrm{cos}2x+2\left(\pi -x\right)\mathrm{sin}x-1}{{\left(\mathrm{cos}2x+\mathrm{cos}x\right)}^{2}}=\underset{t\to 0}{lim}\frac{\mathrm{cos}2t+2t\mathrm{sin}t-1}{{\left(\mathrm{cos}2t-\mathrm{cos}t\right)}^{2}}=$
$=\underset{t\to 0}{lim}\frac{2t\mathrm{sin}t-2{\mathrm{sin}}^{2}t}{{\left(1-\mathrm{cos}t\right)}^{2}{\left(2\mathrm{cos}t+1\right)}^{2}}=\frac{2}{9}\underset{t\to 0}{lim}\frac{t\mathrm{sin}t-{\mathrm{sin}}^{2}t}{4{\mathrm{sin}}^{4}\frac{t}{2}}=$
$=\frac{8}{9}\underset{t\to 0}{lim}\frac{t\mathrm{sin}t-{\mathrm{sin}}^{2}t}{{t}^{4}}=\frac{8}{9}\underset{t\to 0}{lim}\frac{t-\mathrm{sin}t}{{t}^{3}}=$
$=\frac{8}{9}\underset{t\to 0}{lim}\frac{1-\mathrm{cos}\left\{t\right\}}{3{t}^{2}}=\frac{8}{9}\underset{t\to 0}{lim}\frac{2{\mathrm{sin}}^{2}\frac{t}{2}}{3{t}^{2}}=\frac{8}{9}\underset{t\to 0}{lim}\frac{2\cdot \frac{{t}^{2}}{4}}{3{t}^{2}}=\frac{4}{27}$
###### Not exactly what you’re looking for? Charlie Haley
Hint: breaks it like this
$\underset{x\to \pi }{lim}\frac{\mathrm{cos}2x+2\left(\pi -x\right)\mathrm{sin}x-1}{{\left(\mathrm{cos}2x+\mathrm{cos}x\right)}^{2}}$
$=\underset{x\to \pi }{lim}\left(\frac{\mathrm{cos}2x-1}{{\left(\pi -x\right)}^{2}}+2\frac{\mathrm{sin}x}{\left(\pi -x\right)}\right)\cdot \underset{x\to \pi }{lim}{\left(\frac{\pi -x}{\mathrm{cos}2x+\mathrm{cos}x}\right)}^{2}$
Then use the definition of derivative of a function at $x=\pi$