find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using…
find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.\nlim\ncos(x)\n1 - sin(x)\nresources\nread it watch it\nsubmit answer\n7. -/1 points\ndetails my notes ask your teacher practice another\nfind the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.\nlim\n7t\n- 1\nsin(t)
Answer
Explanation:
Step1: Check the form of the limit
When (x = \frac{\pi}{2}), (\cos(\frac{\pi}{2})=0) and (1-\sin(\frac{\pi}{2})=1 - 1=0). So, it is in the (\frac{0}{0}) form.
Step2: Apply L'Hospital's Rule
Differentiate the numerator and denominator. The derivative of (y=\cos(x)) is (y'=-\sin(x)), and the derivative of (y = 1-\sin(x)) is (y'=-\cos(x)). So, (\lim_{x\rightarrow(\frac{\pi}{2})^+}\frac{\cos(x)}{1 - \sin(x)}=\lim_{x\rightarrow(\frac{\pi}{2})^+}\frac{-\sin(x)}{-\cos(x)}=\lim_{x\rightarrow(\frac{\pi}{2})^+}\tan(x))
Step3: Evaluate the new limit
As (x\rightarrow(\frac{\pi}{2})^+), (\tan(x)\rightarrow-\infty)
For (\lim_{t\rightarrow0}\frac{e^{7t}-1}{\sin(t)}):
Step1: Check the form of the limit
When (t = 0), (e^{0}-1=0) and (\sin(0)=0). So, it is in the (\frac{0}{0}) form.
Step2: Apply L'Hospital's Rule
Differentiate the numerator and denominator. The derivative of (y = e^{7t}-1) is (y'=7e^{7t}), and the derivative of (y=\sin(t)) is (y'=\cos(t)) So, (\lim_{t\rightarrow0}\frac{e^{7t}-1}{\sin(t)}=\lim_{t\rightarrow0}\frac{7e^{7t}}{\cos(t)})
Step3: Evaluate the new limit
Substitute (t = 0) into (\frac{7e^{7t}}{\cos(t)}), we get (\frac{7e^{0}}{\cos(0)}=\frac{7\times1}{1}=7)
Answer:
(\lim_{x\rightarrow(\frac{\pi}{2})^+}\frac{\cos(x)}{1 - \sin(x)}=-\infty) (\lim_{t\rightarrow0}\frac{e^{7t}-1}{\sin(t)}=7)