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Limits — JEE Maths practice question

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

Question
If lim⁡x→0[1+x2+f(x)x]1/x2=e7\displaystyle\lim_{x\to 0}\left[1+x^{2}+\dfrac{f(x)}{x}\right]^{1/x^{2}}=e^{7}, then lim⁡x→0f(x)x3\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x^{3}} is equal to.
A77
B44
C55
D66correct
Solution
Step 1: Form is 1∞1^{\infty}. Use the formula:
L=exp⁡[lim⁡x→01x2(x2+f(x)x)]=exp⁡[lim⁡x→0(1+f(x)x3)]=e7L=\exp\left[\lim_{x\to 0}\dfrac{1}{x^{2}}\left(x^{2}+\dfrac{f(x)}{x}\right)\right]=\exp\left[\lim_{x\to 0}\left(1+\dfrac{f(x)}{x^{3}}\right)\right]=e^{7}
 Step 2: Therefore:
1+lim⁡x→0f(x)x3=7  ⇒  lim⁡x→0f(x)x3=61+\lim_{x\to 0}\dfrac{f(x)}{x^{3}}=7\;\Rightarrow\;\lim_{x\to 0}\dfrac{f(x)}{x^{3}}=6
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