We solve the given limit by using L. Hospital’s rule which is,
If $\mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} \frac{{{\text{f}}({\text{x}})}}{{{\text{g}}({\text{x}})}} = \frac{0}{0}$.
Then, $\mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} \frac{{{\text{f}}({\text{x}})}}{{{\text{g}}({\text{x}})}} = \mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} \frac{{{\text{f}}({\text{x}})}}{{{\text{g}}({\text{x}})}}$
So now solving the limit,
$\mathop {\lim }\limits_{x \to 1} \frac{{{x^7} – 2{x^5} + 1}}{{{x^3} – 3{x^2} + 2}}$
Substituting the limit value we get,
$\mathop {\lim }\limits_{x \to 1} \frac{{{x^7} – 2{x^5} + 1}}{{{x^3} – 3{x^2} + 2}} = \frac{0}{0}$
As this is indeterminant so we apply L. Hospital’s rule,
$\mathop {\lim }\limits_{x \to 1} \frac{{{x^7} – 2{x^5} + 1}}{{{x^3} – 3{x^2} + 2}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left( {{x^7} – 2{x^5} + 1} \right)}}{{\frac{d}{{dx}}\left( {{x^3} – 3{x^2} + 2} \right)}}$
$ = \mathop {\lim }\limits_{x \to 1} \frac{{7{x^6} – 10{x^4}}}{{3{x^2} – 6x}}$
$ = \frac{{7 – 10}}{{3 – 6}}$
$ = \frac{{ – 3}}{{ – 3}}$
$ = 1$
Therefore, $\mathop {\lim }\limits_{x \to 1} \frac{{{x^7} – 2{x^5} + 1}}{{{x^3} – 3{x^2} + 2}} = 1$.