Hello, the question is:
limit t-->0+ (cos(t), sin(t), tln(t))
(the limit as t approaches 0 from the right)
Thanks,
-cb
I am working on one approach without using L'Hopital.
(This is not for the original poster. Do not confuse yourself. This is just for fun for other posters).
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Directly by definition.
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We want to show,
$\displaystyle \lim_{t\to 0^+} t\ln t = \lim_{t\to 0^+} t\int_1^t \frac{1}{x} dx = \lim_{t\to 0^+} - t\int_t^1 \frac{1}{x} dx$
Ignore the minus sign for now. Now try to show that $\displaystyle \lim_{t\to 0^+} t\int_t^1 \frac{1}{x} \ dx = 0$.
Let $\displaystyle t$ be small enough so that we are working on the interval $\displaystyle [t,1]$.
Now, $\displaystyle \frac{1}{x} \leq \frac{1}{e^x}$ on $\displaystyle [t,1]$ because $\displaystyle e^x \geq x$ on $\displaystyle (0,1]$. Thus, $\displaystyle \int_t^1 \frac{1}{x} dx \leq \int_1^t \frac{1}{e^x} dx = \frac{1}{e^t}-1$.
This means,
$\displaystyle \left| t\int_t^1 \frac{1}{x} dx \right| \leq \frac{t}{e^t} - t \leq \frac{t}{e^t}$.
But,
$\displaystyle \lim_{t\to 0^+} \frac{t}{e^t} = 0$.
No L'Hopital!