Question

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

Limits and continuity
ANSWERED
asked 2021-01-30
Evaluate the following limits: \(\lim_{y\rightarrow0}\frac{1-\cos(7y)}{2y}\)

Answers (1)

2021-01-31
Given:
\(\lim_{y\rightarrow0}\frac{1-\cos(7y)}{2y}\)
We have to evaluate the given limit
We have,
\(\lim_{y\rightarrow0}\frac{1-\cos(7y)}{2y}\)
When we substitute y=0 in given limit we get,
\(\frac{1-\cos (7\cdot0)}{(2\cdot0)}=\frac{1-1}{0}=\frac{0}{0}\)
We get the indeterminate form if we substitute y=0 then we used L'Hospital's Rule.
L'Hospital's Rule :
The rule tells us that if we have an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) 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:
\(\Rightarrow\lim_{y\rightarrow0}\frac{\frac{d}{dy}(1-cos(7y))}{\frac{d}{dy}(2y)}\)
\(=\lim_{y\rightarrow0}\frac{0-(-7\sin(7y))}{2}\)
\(=\lim_{y\rightarrow0}\frac{7\sin(7y)}{2}\)
\(=\frac{7}{2}\lim_{y\rightarrow0}\sin(7y)\)
\(=0\)
\([\lim_{y\rightarrow0}\sin(7y)=0]\)
Hence, \(\lim_{y\rightarrow0}\frac{1-\cos(7y)}{2y}=0\)
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