GATE CS 2017 Set 1 Q.28 (CS Q.28): Calculus

The value of

\lim_{x \to 1} \frac{x^7 - 2x^5 + 1}{x^3 - 3x^2 + 2}

is 0

is -1

is 1

does not exist

Official answer and solution

Solution

Evaluate the indeterminate limit using L'Hôpital's Rule.

We need to find:

L = \lim_{x \to 1} \frac{x^7 - 2x^5 + 1}{x^3 - 3x^2 + 2}

Evaluating the numerator and denominator at $x = 1$ :

\text{Numerator: } 1^7 - 2(1^5) + 1 = 1 - 2 + 1 = 0

\text{Denominator: } 1^3 - 3(1^2) + 2 = 1 - 3 + 2 = 0

Since the limit is in the indeterminate form $\frac{0}{0}$ , we apply L'Hôpital's Rule.

Differentiating numerator and denominator with respect to $x$ :

\frac{d}{dx}(x^7 - 2x^5 + 1) = 7x^6 - 10x^4

\frac{d}{dx}(x^3 - 3x^2 + 2) = 3x^2 - 6x

Evaluating the limit of the derivatives as $x \to 1$ :

L = \lim_{x \to 1} \frac{7x^6 - 10x^4}{3x^2 - 6x} = \frac{7(1)^6 - 10(1)^4}{3(1)^2 - 6(1)} = \frac{7 - 10}{3 - 6} = \frac{-3}{-3} = 1

Final answer

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