Question

Evaluate the following limits:

${\displaystyle \underset{x-\infty }{lim}\left\{x[\log (x+\text{a})-\log \left(x\right)]\right\}}$

Answer 1

We Know ${\displaystyle \underset{x\to 0}{lim}\frac{\log (1+x)}{x}}$ = 1

${\displaystyle \underset{x-\infty }{lim}\left\{x[\log (x+\text{a})-\log \left(x\right)]\right\}}$

= ${\displaystyle \underset{x>\infty }{lim}x\cdot \log \left(\frac{x+\text{a}}{x}\right)}$

= ${\displaystyle \underset{x\to \infty }{lim}\frac{\log (\frac{x}{x}+\frac{\text{a}}{x})}{\frac{1}{x}}}$

= ${\displaystyle \underset{x\to \infty }{lim}\frac{\log (1+\frac{\text{a}}{x})}{\frac{1}{\text{a}}\times \frac{\text{a}}{x}}}$

= ${\displaystyle \text{a}\underset{x\to \infty }{lim}\frac{\log (1+\frac{\text{a}}{x})}{\frac{\text{a}}{x}}}$ ......(1)

Put y =  ${\displaystyle \frac{\text{a}}{x}}$

When x = ${\displaystyle \infty }$ then y = ${\displaystyle \frac{\text{a}}{\infty }}$ = 0

x → ${\displaystyle \infty }$

⇒ y → 0

(1) ⇒ ${\displaystyle \underset{x-\infty }{lim}\left\{x[\log (x+\text{a})-\log \left(x\right)]\right\}}$

= ${\displaystyle \text{a}\underset{y\to 0}{lim}\frac{\log (1+y)}{y}}$

= a × 1

= a