GATE CS 2017 Set 1 Q28 — Calculus: Limits, Continuity, Differentiability

To find the limit:

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

First, substitute $x = 1$ into the expression to check the form:

• Numerator: $1^7 - 2(1)^5 + 1 = 1 - 2 + 1 = 0$
• Denominator: $1^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0$

Since this results in the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $x$:

$$\lim_{x\rightarrow 1} \frac{\frac{d}{dx}(x^{7}-2x^{5}+1)}{\frac{d}{dx}(x^{3}-3x^{2}+2)} = \lim_{x\rightarrow 1} \frac{7x^6 - 10x^4}{3x^2 - 6x}$$

Now, substitute $x = 1$ into the differentiated expression:

$$\frac{7(1)^6 - 10(1)^4}{3(1)^2 - 6(1)} = \frac{7 - 10}{3 - 6} = \frac{-3}{-3} = 1$$

Thus, the value of the limit is $1$.