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Evaluate the Following Limit: Lim X → 1 X 7 − 2 X 5 + 1 X 3 − 3 X 2 + 2

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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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अध्याय 29: Limits - Exercise 29.3 [पृष्ठ २४]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.3 | Q 34 | पृष्ठ २४

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