# Evaluate limits without L'hôpital's rule

#### idontknow

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) $$\displaystyle \lim_{n\rightarrow \infty } \dfrac{\sqrt{1!+2!+\dotsc + n!}}{n!}\;$$ , $$\displaystyle n\in\mathbb{N}$$.

#### idontknow

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

topsquark

#### greg1313

Forum Staff
$$\displaystyle \lim_{t\rightarrow e } \dfrac{t\ln(t)-e}{t-e}$$

$$\displaystyle t=e+h$$

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

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

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