Question

lim┬(x→0)⁡〖(√(1+x+x^2 )-1)/tan⁡5x 〗

Answer 1

Evaluting the limand directly at x = 0 yields the indeterminate form 0/0, so we have a candidate for L'Hopital's rule. We get

$\displaystyle\lim_{x\to0}\frac{\sqrt{1+x+x^2}-1}{\tan(5x)} = \lim_{x\to0}\frac{\frac{1+2x}{2\sqrt{1+x+x^2}}}{5\sec^2(5x)} = \frac{1+2\times0}{2\sqrt{1+0+0^2}\times5\sec^2(5\times0)} = \boxed{\dfrac{1}{10}}$

If you don't know about L'Hopital's rule, or are otherwise not allowed to use it, you can instead rely on algebraic manipulation and a well-known limit,

$\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=\lim_{x\to0}\frac{ax}{\sin(ax)}=1$

where a ≠ 0. We write

$\displaystyle\frac{\sqrt{1+x+x^2}-1}{\tan(5x)} = \frac{5x}{\sin(5x)}\times\frac{\cos(5x)}5\times\frac{\sqrt{1+x+x^2}-1}x$

The limit of a product is equal to the product of limits:

$\displaystyle\lim_{x\to0}\frac{\sqrt{1+x+x^2}-1}{\tan(5x)} = \left(\lim_{x\to0}\frac{5x}{\sin(5x)}\right)\times\left(\lim_{x\to0}\frac{\cos(5x)}5\right)\times\left(\lim_{x\to0}\frac{\sqrt{1+x+x^2}-1}x\right)$

The first limit is 1 and the second limit is 1/5. For the remaining limit, multiply through the fraction by the conjugate of the numerator:

$\dfrac{\sqrt{1+x+x^2}-1}x\times\dfrac{\sqrt{1+x+x^2}+1}{\sqrt{1+x+x^2}+1} = \dfrac{\left(\sqrt{1+x+x^2}\right)^2-1^2}{x\left(\sqrt{1+x+x^2}+1\right)} \\\cdots= \dfrac{x+x^2}{x\left(\sqrt{1+x+x^2}+1\right)} \\\cdots= \dfrac{1+x}{\sqrt{1+x+x^2}+1}$

The remaining limand is continuous at x = 0, and its limit is

$\displaystyle\lim_{x\to0}\frac{1+x}{\sqrt{1+x+x^2}+1}=\frac{1+0}{\sqrt{1+0+0^2}+1}=\frac12$

and hence the original limit is again 1 × 1/5 × 1/2 = 1/10.