Question

Evaluate :
$i.$ $\underset{x\to 1}{lim}\frac{x^{15}-1}{x^{10}-1}$
$ii.$ $\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}$

Answer 1

$\left(i\right)$ Given: $\underset{x\to 1}{lim}\frac{x^{15}-1}{x^{10}-1}$
$=\frac{(1{)}^{15}-1}{(1{)}^{10}-1}$
$=\frac{1-1}{1-1}$
$=\frac{0}{0}$
It is in the form $\frac{0}{0}$,
Hence, divide numerator and denominator with $(x-1)$

$\Rightarrow \underset{x\to 1}{lim}\frac{x^{15}-1}{x^{10}-1}=\left[\underset{x\to 1}{lim}\frac{x^{15}-1}{x-1}\right]÷\left[\underset{x\to 1}{lim}\frac{x^{10}-1}{x-1}\right]$

$=\left[\underset{x\to 1}{lim}\frac{x^{15}-(1{)}^{15}}{x-1}\right]÷\left[{lim}_{x\to 1}\frac{x^{10}-(1{)}^{10}}{x-1}\right]$
Using, $\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$

$=\left(15\right(1{)}^{15-1})÷\left(10\right(1{)}^{10-1})$
$=15÷10$
$=\frac{15}{10}=\frac{3}{2}$
$\therefore \underset{x\to 1}{lim}\frac{x^{15}-1}{x^{10}-1}=\frac{3}{2}$

$\left(ii\right)$ Given : $\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}$
Substituting $x=0$
$=\frac{\sqrt{1+0}-1}{0}=\frac{1-1}{0}=\frac{0}{0}$
Since it is a $\frac{0}{0}$ form
Take $y=1+x$
$\Rightarrow y-1=x$
As $x\to 0$
$y\to 1+0$
$y\to 1$
So, our limit becomes
$\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}=\underset{y\to 1}{lim}\frac{\sqrt{y}-1}{y-1}$
$=\underset{y\to 1}{lim}\frac{y^{\frac{1}{2}}-1}{y-1}$
$=\underset{y\to 1}{lim}\frac{y^{\frac{1}{2}}-1^{\frac{1}{2}}}{y-1}$
Using, $\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$

​​​​​$=\frac{1}{2}\times 1^{(\frac{1}{2}-1)}$
$=\frac{1}{2}\times 1=\frac{1}{2}$
$\therefore \underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}=\frac{1}{2}$