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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
The value of lim⁡x→1x1/3−x1/4x3−1\displaystyle \lim_{x\to 1}\dfrac{x^{1/3}-x^{1/4}}{x^{3}-1} is
A136\dfrac{1}{36}correct
B−136-\dfrac{1}{36}
C−112-\dfrac{1}{12}
D112\dfrac{1}{12}
Solution
**Step 1: Check the form.** Substituting x=1x=1: numerator =1−1=0=1-1=0, denominator =1−1=0=1-1=0. So the form is 00\dfrac{0}{0}. **Step 2: Recall the standard formula.**
lim⁡x→1xn−1x−1=n(for any real n)\lim_{x\to 1}\dfrac{x^{n}-1}{x-1}=n \quad \text{(for any real } n\text{)}
 **Step 3: Factor the denominator.**
x3−1=(x−1)(x2+x+1)x^{3}-1=(x-1)(x^{2}+x+1)
 **Step 4: Manipulate the numerator by adding and subtracting 1.**
x1/3−x1/4=(x1/3−1)−(x1/4−1)x^{1/3}-x^{1/4}=(x^{1/3}-1)-(x^{1/4}-1)
 **Step 5: Rewrite the limit.**
lim⁡x→1(x1/3−1)−(x1/4−1)(x−1)(x2+x+1)\lim_{x\to 1}\dfrac{(x^{1/3}-1)-(x^{1/4}-1)}{(x-1)(x^{2}+x+1)}
 **Step 6: Split into two fractions.**
=lim⁡x→1 ⁣[x1/3−1x−1−x1/4−1x−1]⋅1x2+x+1=\lim_{x\to 1}\!\left[\dfrac{x^{1/3}-1}{x-1}-\dfrac{x^{1/4}-1}{x-1}\right]\cdot\dfrac{1}{x^{2}+x+1}
 **Step 7: Apply the standard formula to each piece.**
lim⁡x→1x1/3−1x−1=13,lim⁡x→1x1/4−1x−1=14,lim⁡x→11x2+x+1=13\lim_{x\to 1}\dfrac{x^{1/3}-1}{x-1}=\dfrac{1}{3},\quad \lim_{x\to 1}\dfrac{x^{1/4}-1}{x-1}=\dfrac{1}{4},\quad \lim_{x\to 1}\dfrac{1}{x^{2}+x+1}=\dfrac{1}{3}
 **Step 8: Combine.**
(13−14)⋅13=112⋅13=136\left(\dfrac{1}{3}-\dfrac{1}{4}\right)\cdot\dfrac{1}{3}=\dfrac{1}{12}\cdot\dfrac{1}{3}=\dfrac{1}{36}
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