Question

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.
lim
cos(x)
1 - sin(x)
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find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.
lim
7t

• 1

sin(t)

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, \(\lim_{x
ightarrow(\frac{\pi}{2})^+}\frac{\cos(x)}{1 - \sin(x)}=\lim_{x
ightarrow(\frac{\pi}{2})^+}\frac{-\sin(x)}{-\cos(x)}=\lim_{x
ightarrow(\frac{\pi}{2})^+}\tan(x)\)

Step3: Evaluate the new limit

As \(x
ightarrow(\frac{\pi}{2})^+\), \(\tan(x)
ightarrow-\infty\)

For \(\lim_{t
ightarrow0}\frac{e^{7t}-1}{\sin(t)}\):

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, \(\lim_{t
ightarrow0}\frac{e^{7t}-1}{\sin(t)}=\lim_{t
ightarrow0}\frac{7e^{7t}}{\cos(t)}\)

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:

\(\lim_{x
ightarrow(\frac{\pi}{2})^+}\frac{\cos(x)}{1 - \sin(x)}=-\infty\)
\(\lim_{t
ightarrow0}\frac{e^{7t}-1}{\sin(t)}=7\)