NIMCET Previous Year Questions (PYQs) – Limit – Page 1 of 2

Quick DL Method Solution

Given:

$$\underset{x\to 1}{lim}\frac{x^4-1}{x-1}=\underset{x\to k}{lim}\frac{x^3-k^2}{x^2-k^2}$$

LHS using derivative:

$$\underset{x\to 1}{lim}\frac{x^4-1}{x-1}={\frac{d}{dx}(x^4)|}_{x=1}=4x^3{|}_{x=1}=4$$

RHS using DL logic:

$$\underset{x\to k}{lim}\frac{x^3-k^2}{x^2-k^2}\approx \frac{3k^2(x-k)}{2k(x-k)}=\frac{3k}{2}$$

Equating both sides:

$$\frac{3k}{2}=4\Rightarrow k=\frac{8}{3}$$

$$\boxed{{\displaystyle k=\frac{8}{3}}}$$

Evaluate: $$\underset{x\to 0}{lim}\frac{e^x-e^{-x}-2x}{1-\cos x}$$

Step 1: Apply L'Hôpital's Rule (since it's 0/0):

First derivative: $$\frac{e^x+e^{-x}-2}{\sin x}$$

Still 0/0 → Apply L'Hôpital's Rule again: $$\frac{e^x-e^{-x}}{\cos x}$$

Now, $$\underset{x\to 0}{lim}\frac{1-1}{1}=0$$

Final Answer: $$\boxed{{\displaystyle 0}}$$