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without using L hop-ital rule and series expansion , $\displaystyle \lim_{t\rightarrow 0}\bigg(t\cot t+t\ln t\bigg).$

$\displaystyle \lim_{t\rightarrow 0}\bigg(t\frac{\cos t}{\sin t}+\ln t^t\bigg) = 1+\lim_{t\rightarrow 0}\ln(t)^t$

could some help me with this , thanks

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    [How do I calculate $\lim_{x\rightarrow 0} x\ln x$](http://math.stackexchange.com/q/303976) - and you can probably find a few similar questions on the main site. I found this one [from Approach0](https://approach0.xyz/search/?q=%24%5Clim_%7Bt%5Cto0%7D%20t%5Cln%20t%24&p=1). – 2017-01-06

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For $0 < t < 1$

$$ -t\ln t = -2t\ln \sqrt{t}= 2t\int_{\sqrt{t}}^1 \frac{ds}{s} \leqslant 2t\frac{1 - \sqrt{t}}{\sqrt{t}} = 2\sqrt{t}(1 - \sqrt{t})\\ -t\ln t = -2t\ln \sqrt{t}= 2t\int_{\sqrt{t}}^1 \frac{ds}{s} \geqslant 2t(1 - \sqrt{t}) $$

Now use the squeeze theorem to show $t \ln t \to 0$.

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Note that $t \to 0^+$ $$x\to 0 \to \ln(1+x)\sim x$$ $$\displaystyle \lim_{t\rightarrow 0^+}\bigg(t\frac{\cos t}{\sin t}+\ln t^t\bigg) =\\ 1+\lim_{t\rightarrow 0^+}t\ln(t)=\\ 1+\lim_{t\rightarrow 0^+}t\ln(1+(t-1))=\\ 1+\lim_{t\rightarrow 0^+}t\times (t-1)=\\1+0$$