Finding Limits - Math

Card 0 of 4

Question

Evaluate the limit below:

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

Answer

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:

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

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Question

Evaluate the limit below:

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

Answer

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:

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

Compare your answer with the correct one above

Question

Evaluate the limit below:

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

Answer

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:

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

Compare your answer with the correct one above

Question

Evaluate the limit below:

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

Answer

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:

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

Compare your answer with the correct one above

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