37. $lim_{\theta\rightarrow\frac{pi}{2}}\thetasec\theta$\n38. $lim_{y\rightarrow2}ln(sin\frac{pi}{y})$\n39…

37. $lim_{\theta\rightarrow\frac{pi}{2}}\thetasec\theta$\n38. $lim_{y\rightarrow2}ln(sin\frac{pi}{y})$\n39. $lim_{\theta\rightarrow0}\frac{cos\theta - 2}{\theta}$\n40. $lim_{x\rightarrow4.3}\frac{1}{x-lfloor x\rfloor}$\n41. $lim_{x\rightarrow2^{-}}\frac{x - 3}{x - 2}$\n42. $lim_{t\rightarrow0}\frac{sin^{2}t}{t^{3}}$\n43. $lim_{x\rightarrow1^{+}}(\frac{1}{sqrt{x - 1}}-\frac{1}{sqrt{x^{2}-1}})$\n44. $lim_{t\rightarrow e}sqrt{t}(ln t - 1)$\n45. $lim_{x\rightarrow\frac{pi}{2}}\tan x$

Answer

Explanation:

Step1: Recall secant - cosine relation

We know that $\sec\theta=\frac{1}{\cos\theta}$, so $\lim_{\theta\rightarrow\frac{\pi}{2}}\theta\sec\theta=\lim_{\theta\rightarrow\frac{\pi}{2}}\frac{\theta}{\cos\theta}$. As $\theta\rightarrow\frac{\pi}{2}$, $\cos\theta\rightarrow0$ and $\theta\rightarrow\frac{\pi}{2}$, the limit does not exist.

Step2: Evaluate $\lim_{y\rightarrow2}\ln(\sin\frac{\pi}{y})$

First, find the value of $\sin\frac{\pi}{y}$ when $y\rightarrow2$. When $y = 2$, $\sin\frac{\pi}{y}=\sin\frac{\pi}{2}=1$. Then $\lim_{y\rightarrow2}\ln(\sin\frac{\pi}{y})=\ln(\lim_{y\rightarrow2}\sin\frac{\pi}{y})=\ln(1) = 0$.

Step3: Evaluate $\lim_{\theta\rightarrow0}\frac{\cos\theta - 2}{\theta}$

We know that $\cos\theta=1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}-\cdots$. So $\cos\theta - 2=-1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}-\cdots$. Then $\lim_{\theta\rightarrow0}\frac{\cos\theta - 2}{\theta}=\lim_{\theta\rightarrow0}\frac{-1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}-\cdots}{\theta}$, and this limit does not exist (the left - hand limit and right - hand limit are not equal).

Step4: Evaluate $\lim_{x\rightarrow4.3}\frac{1}{x-\lfloor x\rfloor}$

Since $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$, when $x\rightarrow4.3$, $\lfloor x\rfloor = 4$, and $x-\lfloor x\rfloor\rightarrow0.3$. So $\lim_{x\rightarrow4.3}\frac{1}{x - \lfloor x\rfloor}=\frac{1}{0.3}=\frac{10}{3}$.