Evaluate this trig limit without L'Hospital: $\underset{x\to \text{}0}{lim}\phantom{\rule{thickmathspace}{0ex}}x\mathrm{tan}\left(\frac{\pi (1-x)}{2}\right)$

Cory Patrick
2022-07-01
Answered

Evaluate this trig limit without L'Hospital: $\underset{x\to \text{}0}{lim}\phantom{\rule{thickmathspace}{0ex}}x\mathrm{tan}\left(\frac{\pi (1-x)}{2}\right)$

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Mateo Barajas

Answered 2022-07-02
Author has **13** answers

Using some trig identities, we have

$\begin{array}{rl}x\mathrm{tan}(\frac{\pi}{2}-\frac{\pi x}{2})& =x\mathrm{cot}\left(\frac{\pi x}{2}\right)\\ & =x\frac{\mathrm{cos}\left(\frac{\pi x}{2}\right)}{\mathrm{sin}\left(\frac{\pi x}{2}\right)}\\ & =\frac{2}{\pi}\mathrm{cos}\left(\frac{\pi x}{2}\right){\left(\frac{\mathrm{sin}\left(\frac{\pi x}{2}\right)}{\frac{\pi x}{2}}\right)}^{-1}\end{array}$

And recall how to do limits of the form $\frac{\mathrm{sin}(u)}{u}.$

$\begin{array}{rl}x\mathrm{tan}(\frac{\pi}{2}-\frac{\pi x}{2})& =x\mathrm{cot}\left(\frac{\pi x}{2}\right)\\ & =x\frac{\mathrm{cos}\left(\frac{\pi x}{2}\right)}{\mathrm{sin}\left(\frac{\pi x}{2}\right)}\\ & =\frac{2}{\pi}\mathrm{cos}\left(\frac{\pi x}{2}\right){\left(\frac{\mathrm{sin}\left(\frac{\pi x}{2}\right)}{\frac{\pi x}{2}}\right)}^{-1}\end{array}$

And recall how to do limits of the form $\frac{\mathrm{sin}(u)}{u}.$

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