Finding Limits Flashcards - High School Math

will approach $1^{\infty }$ when approaches $\frac{\pi }{2}$ , so $\lim_{x\rightarrow \frac{\pi }{2}} (lny)$ will be of type $0\cdot \infty$ as shown below:

$\lim_{x\rightarrow \frac{\pi }{2}} lny = \lim_{x\rightarrow \frac{\pi }{2}} ln (sinx-cosx)^{tanx}=\lim_{x\rightarrow \frac{\pi }{2}} tanx\cdot ln(sinx-cosx) = \infty \cdot 0$

So, we can apply the L’ Hospital's Rule:

$\lim_{x\rightarrow \frac{\pi }{2}} lny = \lim_{x\rightarrow \frac{\pi }{2}} \frac{sinx \cdot ln(sinx-cosx)}{cosx}=$

$\lim_{x\rightarrow \frac{\pi }{2}} \frac{cosx\cdot ln(sinx-cosx)+sinx\cdot (\frac{cosx+sinx}{sinx-cosx})}{-sinx}=\frac{0+1}{-1}=-1$

since:

$\lim_{x\rightarrow \frac{\pi }{2}} lny=-1$

hence:

$\lim_{x\rightarrow \frac{\pi }{2}} y=\lim_{x\rightarrow \frac{\pi }{2}} e^{lny}=e^{-1}=\frac{1}{e}$

Finding Limits - High School Math

Evaluate the limit below:

$y=\lim_{x \to\frac{\pi }2{} }(sinx-cosx)^{tanx}$

Finding Limits - High School Math

Evaluate the limit below:

will approach when approaches , so will be of type as shown below: So, we can apply the L’ Hospital's Rule: since: hence:

Evaluate the limit below:

$y=\lim_{x \to\frac{\pi }2{} }(sinx-cosx)^{tanx}$