Evaluate limits without L'hôpital's rule

Dec 2015
859
115
Earth
Evaluate :
(a) \(\displaystyle L=\lim_{t\rightarrow 0 } \sqrt{t}\ln^{2}(t)\).

(b) \(\displaystyle L=\lim_{t\rightarrow 0 }\dfrac{1-\sqrt[\displaystyle n]{\cos(t)\cos(2t)\cdot \dotsc \cdot \cos(nt)}}{t^2 }\; \) , \(\displaystyle n\in\mathbb{N}\).

(c) \(\displaystyle \lim_{t\rightarrow e } \dfrac{t\ln(t)-e}{t-e} \; \) , where \(\displaystyle \ln(e)=1\).

(d) \(\displaystyle \lim_{n\rightarrow \infty } \dfrac{\sqrt{1!+2!+\dotsc + n!}}{n!}\; \) , \(\displaystyle n\in\mathbb{N}\).
 
Dec 2015
859
115
Earth
Are the results correct ?

(a) \(\displaystyle L=\lim_{t=e^{-u} , u\rightarrow \infty} \dfrac{u^2 }{e^{u/2}}=4\lim_{u\rightarrow \infty}\dfrac{u^2 }{e^u} =4\lim_{u\rightarrow \infty}\dfrac{u^2 }{\sum_{k=0}^{\infty} \dfrac{u^k }{k!}}=0\).

(d) \(\displaystyle L=\sqrt{\lim_{n\rightarrow \infty }\dfrac{1!+2!+...+n!}{n!^2 }}= \sqrt{\lim_{n\rightarrow \infty }\dfrac{n!(n+1)}{n!^2 (n^2 +2n ) }}=0\).

(b) \(\displaystyle L=\lim_{t\rightarrow 0 } \dfrac{-\ln\cos(t)-\ln\cos(2t)-...-\ln\cos(nt) }{t^2 }=\dfrac{1^2 + 2^2 +...+ n^2 }{2}=n(n+1)(2n+1)/12\).
 
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greg1313

Forum Staff
Oct 2008
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London, Ontario, Canada - The Forest City
\(\displaystyle \lim_{t\rightarrow e } \dfrac{t\ln(t)-e}{t-e}\)

\(\displaystyle t=e+h\)

\(\displaystyle f(t)=t\ln(t)\)

\(\displaystyle \lim_{h\to0}\frac{(e+h)\ln(e+h)-e}{h}=f'(e)=2\)
 
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