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प्रश्न
Evaluate the following limit:
\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]
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उत्तर
When x = 1, the expression \[\frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]assumes the form \[\frac{0}{0}\] So, (x − 1) is a factor of numerator and denominator.
Using long division method, we get
\[x^7 - 2 x^5 + 1 = \left( x - 1 \right)\left( x^6 + x^5 - x^4 - x^3 - x^2 - x - 1 \right)\] and \[x^3 - 3 x^2 + 2 = \left( x - 1 \right)\left( x^2 - 2x - 2 \right)\]
\[\therefore \lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]
\[ = \lim_{x \to 1} \frac{\left( x - 1 \right)\left( x^6 + x^5 - x^4 - x^3 - x^2 - x - 1 \right)}{\left( x - 1 \right)\left( x^2 - 2x - 2 \right)}\]
\[ = \lim_{x \to 1} \frac{x^6 + x^5 - x^4 - x^3 - x^2 - x - 1}{x^2 - 2x - 2}\]
\[ = \frac{1 + 1 - 1 - 1 - 1 - 1 - 1}{1 - 2 - 2}\]
\[ = \frac{- 3}{- 3}\]
\[ = 1\]
\[\]
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