Question

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

Limits and continuity
Evaluate the following limits: $$\lim_{y\rightarrow0}\frac{1-\cos(7y)}{2y}$$

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, $$\lim_{y\rightarrow0}\frac{1-\cos(7y)}{2y}=0$$