veksetz

2022-01-16

Evaluate the following in terms of $\alpha =\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}$:
$\underset{x\to 0}{lim}\frac{{\mathrm{tan}}^{2}\left(x\right)+2x}{x+{x}^{2}}$

Lakisha Archer

Given,
$\underset{x\to 0}{lim}\frac{{\mathrm{tan}}^{2}\left(x\right)+2x}{x+{x}^{2}}$
Now taking x common from numerator and denominator,
we get,
$\underset{x\to 0}{lim}\frac{{\mathrm{tan}}^{2}\left(x\right)+2x}{x+{x}^{2}}=\frac{\mathrm{tan}\left(x\right)\frac{\mathrm{tan}x}{x}+2}{1+x}$
as we know, $\underset{x\to 0}{lim}\frac{\mathrm{tan}x}{x}=1$
So,
$\underset{x\to 0}{lim}\frac{\mathrm{tan}\left(x\right)\frac{\mathrm{tan}x}{x}+2}{1+x}=\frac{0\cdot 1+2}{1+0}=2$

RizerMix

You could write $\frac{{\mathrm{tan}}^{2}x+2x}{x+{x}^{2}}=\frac{{\mathrm{tan}}^{2}+2x}{x\left(x+1\right)}=\frac{\frac{\mathrm{sin}x}{x}\cdot \frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}+2}{x+1}$ and use limit laws to get the limit $\frac{\alpha \cdot 0+2}{1}$

alenahelenash

Since $\underset{x\to 0}{lim}\mathrm{cos}x=1$ we have $\underset{x\to 0}{lim}\frac{\mathrm{tan}x}{x}=\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=\alpha$ , i.e. $\mathrm{tan}x=\alpha x+o\left(x\right)$ So ${\mathrm{tan}}^{2}x={\alpha }^{2}{x}^{2}+o\left({x}^{2}\right)$ and ${\mathrm{tan}}^{2}x+2x=2x+o\left(x\right)=x\left(2+o\left(1\right)\right)$ Then $\frac{{\mathrm{tan}}^{2}x+2x}{x\left(1+x\right)}=\frac{2+o\left(1\right)}{1+x}$ and the limit as $x\to 0$ is 2