# Evaluate the following limits: lim_{yrightarrow0}frac{1-cos(7y)}{2y}

Evaluate the following limits: $\underset{y\to 0}{lim}\frac{1-\mathrm{cos}\left(7y\right)}{2y}$
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yagombyeR
Given:
$\underset{y\to 0}{lim}\frac{1-\mathrm{cos}\left(7y\right)}{2y}$
We have to evaluate the given limit
We have,
$\underset{y\to 0}{lim}\frac{1-\mathrm{cos}\left(7y\right)}{2y}$
When we substitute y=0 in given limit we get,
$\frac{1-\mathrm{cos}\left(7\cdot 0\right)}{\left(2\cdot 0\right)}=\frac{1-1}{0}=\frac{0}{0}$
We get the indeterminate form if we substitute y=0 then we used LHospitals Rule.
LHospitals Rule :
The rule tells us that if we have an indeterminate form $\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
Now differentiate the numerator and differentiate the denominator with respect to y:
$⇒\underset{y\to 0}{lim}\frac{\frac{d}{dy}\left(1-cos\left(7y\right)\right)}{\frac{d}{dy}\left(2y\right)}$
$=\underset{y\to 0}{lim}\frac{0-\left(-7\mathrm{sin}\left(7y\right)\right)}{2}$
$=\underset{y\to 0}{lim}\frac{7\mathrm{sin}\left(7y\right)}{2}$
$=\frac{7}{2}\underset{y\to 0}{lim}\mathrm{sin}\left(7y\right)$
$=0$
$\left[\underset{y\to 0}{lim}\mathrm{sin}\left(7y\right)=0\right]$
Hence, $\underset{y\to 0}{lim}\frac{1-\mathrm{cos}\left(7y\right)}{2y}=0$