Consider the limit:

\(\lim_{t\rightarrow0}\frac{5t^2}{\cos t-1}\)

If we try to evaluate the limit by direct substitution we get 0/0.

So, evaluate using l’Hôpital’s Rule

\(\lim_{t\rightarrow0}\frac{5t^2}{\cos t-1}=5\lim_{t\rightarrow0}(\frac{\frac{d}{dt}(t^2)}{\frac{d}{dt}(\cos t-1)})\)

\(=5\lim_{t\rightarrow0}(\frac{2t}{-\sin t})\)

We het 0/0 form, so apply again L'Hospital Rule

\(5\lim_{t\rightarrow0}(\frac{\frac{d}{dt}(2t)}{\frac{d}{dt}(-\sin t)})=5\lim_{t\rightarrow0}(\frac{2}{-\cos t})\)

\(=5(\frac{2}{-\cos0})\)

\(=5(-\frac{2}{-1})\)

\(=-10\)

\(\lim_{t\rightarrow0}\frac{5t^2}{\cos t-1}\)

If we try to evaluate the limit by direct substitution we get 0/0.

So, evaluate using l’Hôpital’s Rule

\(\lim_{t\rightarrow0}\frac{5t^2}{\cos t-1}=5\lim_{t\rightarrow0}(\frac{\frac{d}{dt}(t^2)}{\frac{d}{dt}(\cos t-1)})\)

\(=5\lim_{t\rightarrow0}(\frac{2t}{-\sin t})\)

We het 0/0 form, so apply again L'Hospital Rule

\(5\lim_{t\rightarrow0}(\frac{\frac{d}{dt}(2t)}{\frac{d}{dt}(-\sin t)})=5\lim_{t\rightarrow0}(\frac{2}{-\cos t})\)

\(=5(\frac{2}{-\cos0})\)

\(=5(-\frac{2}{-1})\)

\(=-10\)