# How to evaluate each limit below. (a) lim x+2/x^2+x+1 x→∞ (b) lim (1/√x−2 − 4/x−4)? x->4

a)

$\setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{x + 2}{{x}^{2} + x + 1}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{\frac{x}{x} ^ 2 + \frac{2}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2 + \frac{1}{x} ^ 2}$

$= \setminus {\lim}_{x \setminus \to \setminus \infty} \setminus \frac{\frac{1}{x} + \frac{2}{x} ^ 2}{1 + \frac{1}{x} + \frac{1}{x} ^ 2}$

$= \setminus \frac{0 + 0}{1 + 0 + 0}$

$= 0$

b)

${\lim}_{x \setminus \to 4} \setminus \frac{1}{\setminus \sqrt{x} - 2} - \frac{4}{x - 4}$

$= {\lim}_{x \setminus \to 4} \setminus \frac{x - 4 - 4 \left(\setminus \sqrt{x} - 2\right)}{\left(\setminus \sqrt{x} - 2\right) \left(x - 4\right)}$

$= {\lim}_{x \setminus \to 4} \setminus \frac{x - 4 - 4 \setminus \sqrt{x} + 8}{\left(\setminus \sqrt{x} - 2\right) \left(x - 4\right)}$

$= {\lim}_{x \setminus \to 4} \setminus \frac{x - 4 \setminus \sqrt{x} + 4}{\left(\setminus \sqrt{x} - 2\right) \left(x - 4\right)}$

$= {\lim}_{x \setminus \to 4} \setminus \frac{{\left(\setminus \sqrt{x} - 2\right)}^{2}}{\left(\setminus \sqrt{x} - 2\right) \left(x - 4\right)}$

$= {\lim}_{x \setminus \to 4} \setminus \frac{\setminus \sqrt{x} - 2}{x - 4}$

Applying L'Hopital's rule for $\frac{0}{0}$ form,

$= {\lim}_{x \setminus \to 4} \setminus \frac{\frac{d}{\mathrm{dx}} \left(\setminus \sqrt{x} - 2\right)}{\frac{d}{\mathrm{dx}} \left(x - 4\right)}$

$= {\lim}_{x \setminus \to 4} \setminus \frac{\frac{1}{2 \setminus \sqrt{x}}}{1}$

$= \frac{1}{2} {\lim}_{x \setminus \to 4} \frac{1}{\setminus \sqrt{x}}$

$= \frac{1}{2} \setminus \cdot \frac{1}{2}$

$= \frac{1}{4}$

Jul 17, 2018

$\frac{1}{4}$.

#### Explanation:

Part (b) can be solved without using L'Hospital's Rule :

Observe that, $x - 4 = {\left(\sqrt{x}\right)}^{2} - {2}^{2} = \left(\sqrt{x} + 2\right) \left(\sqrt{x} - 2\right)$.

$\text{The Reqd. Lim.} = {\lim}_{x \to 4} \left\{\frac{1}{\sqrt{x} - 2} - \frac{4}{x - 4}\right\}$,

$= {\lim}_{x \to 4} \left[\frac{1}{\sqrt{x} - 2} - \frac{4}{\left(\sqrt{x} + 2\right) \left(\sqrt{x} - 2\right)}\right]$,

$= {\lim}_{x \to 4} \left[\frac{\left(\sqrt{x} + 2\right) - 4}{\left(\sqrt{x} + 2\right) \left(\sqrt{x} - 2\right)}\right]$,

$= {\lim}_{x \to 4} \frac{\sqrt{x} - 2}{\left(\sqrt{x} + 2\right) \left(\sqrt{x} - 2\right)}$,

$= {\lim}_{x \to 4} \frac{1}{\sqrt{x} + 2}$,

$= \frac{1}{\sqrt{4} + 2}$,

$= \frac{1}{4}$, as Respected Harish Chandra Rajpoot has readily

derived!