# Evaluate the following limits. $$\displaystyle\lim_{{\theta\rightarrow{0}}}{\frac{{{\sec{\theta}}-{1}}}{{\theta}}}$$

Marzadri9lyy 2022-03-28 Answered
Evaluate the following limits.
$\underset{\theta \to 0}{lim}\frac{\mathrm{sec}\theta -1}{\theta }$
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## Answers (1)

cineworld93uowb
Answered 2022-03-29 Author has 16 answers
We have to find the limit:
$\underset{\theta \to 0}{lim}\frac{\mathrm{sec}\theta -1}{\theta }$
We know the identity,
$\mathrm{cos}\theta \mathrm{sec}\theta =1$
$⇒\mathrm{sec}\theta =\frac{1}{\mathrm{cos}\theta }$
Substituting above value in the limit,
$\underset{\theta \to 0}{lim}\frac{\mathrm{sec}\theta -1}{\theta }=\frac{\frac{1}{\mathrm{cos}\theta }-1}{\theta }$
$=\underset{\theta \to 0}{lim}\frac{\frac{1-\mathrm{cos}\theta }{\mathrm{cos}\theta }}{\theta }$
$=\underset{\theta \to 0}{lim}\frac{1-\mathrm{cos}\theta }{\theta \mathrm{cos}\theta }$
We have identity,
${\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=\mathrm{cos}2x$
$⇒1-{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{2}x=\mathrm{cos}2x$
$1-\mathrm{cos}2x=2{\mathrm{sin}}^{2}\frac{\theta }{2}$
then for $1-\mathrm{cos}2x=2{\mathrm{sin}}^{2}\frac{\theta }{2}$
Therefore further limit can be simplified as follows,
$\underset{\theta \to 0}{lim}\frac{1-\mathrm{cos}\theta }{\theta \mathrm{cos}\theta }=\underset{\theta \to 0}{lim}\frac{2{\mathrm{sin}}^{2}\left(\frac{\theta }{2}\right)}{\theta \mathrm{cos}\theta }$
Multiplying and dividing by ${\left(\frac{\theta }{2}\right)}^{2}$, we get
$\underset{\theta \to 0}{lim}\frac{2{\mathrm{sin}}^{2}\left(\frac{\theta }{2}\right)}{\theta \mathrm{cos}\theta }×\frac{{\left(\frac{\theta }{2}\right)}^{2}}{{\left(\frac{\theta }{2}\right)}^{2}}=\underset{\theta \to 0}{lim}\frac{2}{\theta \mathrm{cos}\theta }×\frac{{\theta }^{2}}{4}×\frac{{\mathrm{sin}}^{2}\left(\frac{\theta }{2}\right)}{{\left(\frac{\theta }{2}\right)}^{2}}$
$=\underset{\theta \to 0}{lim}\frac{2}{\mathrm{cos}\theta }×\frac{\theta }{4}×1$
$=\frac{2}{\mathrm{cos}0}×\frac{0}{4}$
$=0$
Hence, value of limit is 0.
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