 # Evaluate the following limits. lim_{thetarightarrow0}frac{sectheta-1}{theta} cistG 2020-11-23 Answered
Evaluate the following limits.
$\underset{\theta \to 0}{lim}\frac{\mathrm{sec}\theta -1}{\theta }$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Tuthornt
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.