How do we find those limits?

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December 14th 2010, 05:26 AM
RCola

How do we find those limits?

What is the steps we need to do in order to solve those limits?

1. Limit [x->1] (x^4-x^3+x^2-3x+2) / (x^3-x^2-x+1) = 2

2. Limit [x-> PI/4] (1-tan^2(x)) / (sqrt(2)*cos(x)-1) = 4

3. Limit [x->PI/2] ( (2x+1) / (x) )^[e^[-x]] = (2+2/PI)^[(-e*PI)/(2)]

4. Limit [x->0] (sqrt(1+x)-x)^[1/x] = 1/sqrt(e)

5. Limit [x->1] (sqrt(x)+sqrt(x-1)-1) / (sqrt( (x^2)-1) ) = 1/sqrt(2)
December 14th 2010, 05:33 AM
Prove It

$\displaystyle \lim_{x \to 1}\frac{x^4 - x^3 + x^2 - 3x + 2}{x^3 - x^2 - x + 1} = \lim_{x \to 1}\frac{(x-1)^2(x^2 + x + 2)}{(x-1)^2(x+1)}$

$\displaystyle = \lim_{x \to 1}\frac{x^2+x+2}{x+1}$

$\displaystyle = \frac{1^2+1+2}{1+1}$

$\displaystyle = \frac{4}{2}$

$\displaystyle = 2$ .
December 14th 2010, 05:57 AM
Prove It

2. $\displaystyle \lim_{x \to \frac{\pi}{4}}\frac{1 - \tan^2{x}}{\sqrt{2}\,\cos{x} - 1} = \lim_{x \to \frac{\pi}{4}}\frac{-2\tan{x}\sec^2{x}}{\sqrt{2}\,\sin{x}}$ by L'Hospital's Rule

$\displaystyle = \frac{-2\tan{\frac{\pi}{4}}\sec^2{\frac{\pi}{4}}}{\sqrt{2 }\,\sin{\frac{\pi}{4}}}$

$\displaystyle = \frac{-2(1)(\sqrt{2})^2}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \right)}$

$\displaystyle = -4$ .
December 14th 2010, 06:02 AM
Prove It

3. $\displaystyle \lim_{x \to \frac{\pi}{2}}\left(\frac{2x+1}{x}\right)^{e^{-x}} = \left[\frac{2\left(\frac{\pi}{2}\right) + 1}{\frac{\pi}{2}}\right]^{e^{-\frac{\pi}{2}}}$

$\displaystyle = \left(\frac{\pi + 1}{\frac{\pi}{2}}\right)^{e^{-\frac{\pi}{2}}}$

$\displaystyle = \left(\frac{2\pi + 2}{\pi}\right)^{e^{-\frac{\pi}{2}}}$

$\displaystyle = \left(2 + \frac{2}{\pi}\right)^{e^{-\frac{\pi}{2}}}$
December 14th 2010, 06:24 AM
Prove It

4. $\displaystyle \lim_{x\to 0}\left(\sqrt{1+x}-x\right)^{\frac{1}{x}} = \lim_{x \to 0}e^{\ln{[(\sqrt{1+x}-x)^{\frac{1}{x}}]}}$

$\displaystyle = \lim_{x\to 0}e^{\frac{1}{x}\ln{(\sqrt{1+x}-x)}}$

$\displaystyle = \lim_{x \to 0}e^{\frac{\ln{(\sqrt{1+x}-x)}}{x}}$

$\displaystyle = e^{\lim_{x \to 0}\frac{\ln{(\sqrt{1+x}-x)}}{x}}$

$\displaystyle = e^{\lim_{x \to 0}\frac{\frac{2-\frac{1}{\sqrt{1+x}}}{2x-2\sqrt{1+x}}}{1}}$ by L'Hospital's Rule

$\displaystyle = e^{\lim_{x \to 0}\frac{2 - \frac{1}{\sqrt{1+x}}}{2x - 2\sqrt{1+x}}}$

$\displaystyle = e^{\frac{2 - \frac{1}{\sqrt{1+0}}}{2(0) - 2\sqrt{1+0}}}$

$\displaystyle = e^{-\frac{1}{2}}$

$\displaystyle = \frac{1}{\sqrt{e}}$ .