Question

If \[ \lim_{x\to0}\left(\frac{p\sin2x+1-\cos2x}{x+\tan x}\right)=1, \] then the value of $p$ is:

• A. $\dfrac{1}{2}$
• B. $-1$
• C. $2$
• D. $1$

Hint:

Whenever direct substitution gives $\dfrac{0}{0}$, think immediately about L'Hôpital's Rule or standard trigonometric limits.

Answer 1

The Correct Option is D

Concept: The given limit produces the indeterminate form: \[ \frac{0}{0} \] Hence, we can apply L'Hôpital's Rule. According to L'Hôpital's Rule: \[ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} \] provided both numerator and denominator approach $0$.

Step 1: Checking the indeterminate form.
Substitute $x=0$. Numerator: \[ p\sin0+1-\cos0 \] \[ =0+1-1=0 \] Denominator: \[ 0+\tan0=0 \] Thus the expression is of type: \[ \frac{0}{0} \]

Step 2: Differentiating numerator and denominator.
Differentiate numerator: \[ \frac{d}{dx}(p\sin2x+1-\cos2x) \] \[ =2p\cos2x+2\sin2x \] Differentiate denominator: \[ \frac{d}{dx}(x+\tan x) \] \[ =1+\sec^2x \]

Step 3: Applying the limit again.
Now: \[ \lim_{x\to0} \frac{2p\cos2x+2\sin2x}{1+\sec^2x} =1 \] Substitute $x=0$: \[ \frac{2p(1)+2(0)}{1+1}=1 \] \[ \frac{2p}{2}=1 \] \[ p=1 \] Hence, \[ \boxed{p=1} \]