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)(√(1+x+x^2)-1)/(\tan(5x)) = \lim_(x\to0)((1+2x)/(2√(1+x+x^2)))/(5\sec^2(5x)) = (1+2*0)/(2√(1+0+0^2)*5\sec^2(5*0)) = \boxed{(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)(\sin(ax))/(ax)=\lim_(x\to0)(ax)/(\sin(ax))=1`

where a ≠ 0. We write

`\displaystyle(√(1+x+x^2)-1)/(\tan(5x)) = (5x)/(\sin(5x))*\frac{\cos(5x)}5*\frac{√(1+x+x^2)-1}x`

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

`\displaystyle\lim_(x\to0)(√(1+x+x^2)-1)/(\tan(5x)) = \left(\lim_(x\to0)(5x)/(\sin(5x))\right)*\left(\lim_(x\to0)\frac{\cos(5x)}5\right)*\left(\lim_(x\to0)\frac{√(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:

`\frac{√(1+x+x^2)-1}x*(√(1+x+x^2)+1)/(√(1+x+x^2)+1) = (\left(√(1+x+x^2)\right)^2-1^2)/(x\left(√(1+x+x^2)+1\right)) \\\cdots= (x+x^2)/(x\left(√(1+x+x^2)+1\right)) \\\cdots= (1+x)/(√(1+x+x^2)+1)`

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

`\displaystyle\lim_(x\to0)(1+x)/(√(1+x+x^2)+1)=(1+0)/(√(1+0+0^2)+1)=\frac12`

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