Evaluate $\lim_{x \to -3} \dfrac{x^3 + 27}{x^5 + 243}$.

Evaluate limx → -3 (x³ + 27)/(x⁵ + 243).

NCERT Exemplar · Q12

\lim_{x \to -3} \dfrac{x^3 + 27}{x^5 + 243}

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Direct substitution gives $\frac{0}{0}$ , so we use the standard limit $\lim_{x\to a}\dfrac{x^n - a^n}{x - a} = n\,a^{n-1}$ on the numerator and denominator separately. The value is $\dfrac{1}{15}$ .

Step 1 — Check the form

Substitute $x = -3$ :

\frac{(-3)^3 + 27}{(-3)^5 + 243} = \frac{-27 + 27}{-243 + 243} = \frac{0}{0},

an indeterminate form, so we must simplify before substituting.

Step 2 — Write the top and bottom as differences of powers

Since $27 = 3^3 = -(-3)^3$ and $243 = 3^5 = -(-3)^5$ :

x^3 + 27 = x^3 - (-3)^3, \qquad x^5 + 243 = x^5 - (-3)^5.

Divide numerator and denominator by the common vanishing factor $\big(x-(-3)\big)$ :

\frac{x^3 + 27}{x^5 + 243} = \frac{\dfrac{x^3 - (-3)^3}{x - (-3)}}{\dfrac{x^5 - (-3)^5}{x - (-3)}}.

Step 3 — Apply the standard limit …

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Q.Evaluate $\lim_{x \to -3} \dfrac{x^3 + 27}{x^5 + 243}$ .

Concept understanding — Standard Limit of (xⁿ − aⁿ)/(x − a)

Step 1 — Check the form

Step 2 — Write the top and bottom as differences of powers

Step 3 — Apply the standard limit …

Q.Evaluate lim⁡x→−3x3+27x5+243\lim_{x \to -3} \dfrac{x^3 + 27}{x^5 + 243}limx→−3​x5+243x3+27​.

Step 1 — Check the form

Step 2 — Write the top and bottom as differences of powers

Step 3 — Apply the standard limit …

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Q.Evaluate $\lim_{x \to -3} \dfrac{x^3 + 27}{x^5 + 243}$ .

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