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Find 
lim_(theta rarr(pi)/(4))(cos(2theta))/(sqrt2cos(theta)-1).
Choose 1 answer:
(A) 2
(B) 
(1)/(2)
(C) 
sqrt2
(D) The limit doesn't exist

Find limθπ4cos(2θ)2cos(θ)1 \lim _{\theta \rightarrow \frac{\pi}{4}} \frac{\cos (2 \theta)}{\sqrt{2} \cos (\theta)-1} .\newlineChoose 11 answer:\newline(A) 22\newline(B) 12 \frac{1}{2} \newline(C) 2 \sqrt{2} \newline(D) The limit doesn't exist

Full solution

Q. Find limθπ4cos(2θ)2cos(θ)1 \lim _{\theta \rightarrow \frac{\pi}{4}} \frac{\cos (2 \theta)}{\sqrt{2} \cos (\theta)-1} .\newlineChoose 11 answer:\newline(A) 22\newline(B) 12 \frac{1}{2} \newline(C) 2 \sqrt{2} \newline(D) The limit doesn't exist
  1. Identify Limit: Identify the limit that needs to be evaluated. We need to find the limit of the function cos(2θ)2cos(θ)1\frac{\cos(2\theta)}{\sqrt{2}\cos(\theta)-1} as θ\theta approaches π4\frac{\pi}{4}.
  2. Check Direct Substitution: Substitute the value of θ\theta into the function to check for direct substitution.\newlineIf we substitute θ=π4\theta = \frac{\pi}{4} into the function, we get cos(2×π4)2×cos(π4)1\frac{\cos(2 \times \frac{\pi}{4})}{\sqrt{2} \times \cos(\frac{\pi}{4}) - 1}, which simplifies to cos(π2)2×(22)1\frac{\cos(\frac{\pi}{2})}{\sqrt{2} \times (\frac{\sqrt{2}}{2}) - 1}. Since cos(π2)=0\cos(\frac{\pi}{2}) = 0, the numerator becomes 00. However, the denominator also becomes (11)=0(1 - 1) = 0, which means we have an indeterminate form 00\frac{0}{0}. Therefore, direct substitution does not work.
  3. Simplify Expression: Simplify the expression and try to eliminate the indeterminate form.\newlineWe can use trigonometric identities to simplify the expression. The double angle formula for cosine is cos(2θ)=cos2(θ)sin2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta), which can also be written as cos(2θ)=2cos2(θ)1\cos(2\theta) = 2\cos^2(\theta) - 1 since sin2(θ)=1cos2(θ)\sin^2(\theta) = 1 - \cos^2(\theta).
  4. Apply Double Angle Identity: Apply the double angle identity to the numerator.\newlineUsing the identity from Step 33, we can rewrite the numerator as 2cos2(θ)12\cos^2(\theta) - 1. So the function becomes (2cos2(θ)1)/(2cos(θ)1)(2\cos^2(\theta) - 1)/(\sqrt{2}\cos(\theta) - 1).
  5. Factor Numerator: Factor the numerator and look for common factors.\newlineThe numerator 2cos2(θ)12\cos^2(\theta) - 1 can be factored as 2(cos2(θ)12)2(\cos^2(\theta) - \frac{1}{2}). We recognize that cos2(θ)12\cos^2(\theta) - \frac{1}{2} is the same as cos2(θ)(22)2\cos^2(\theta) - (\frac{\sqrt{2}}{2})^2, which is a difference of squares and can be factored further.
  6. Factor Difference of Squares: Factor the difference of squares in the numerator.\newlineWe can write cos2(θ)(2/2)2\cos^2(\theta) - (\sqrt{2}/2)^2 as (cos(θ)2/2)(cos(θ)+2/2)(\cos(\theta) - \sqrt{2}/2)(\cos(\theta) + \sqrt{2}/2). Now the function is (2(cos(θ)2/2)(cos(θ)+2/2))/(2cos(θ)1)(2(\cos(\theta) - \sqrt{2}/2)(\cos(\theta) + \sqrt{2}/2))/(\sqrt{2}\cos(\theta) - 1).
  7. Cancel Common Factor: Cancel out the common factor.\newlineWe notice that (cos(θ)22)(\cos(\theta) - \frac{\sqrt{2}}{2}) is a common factor in both the numerator and the denominator. We can cancel this factor out, leaving us with 2(cos(θ)+22)2\frac{2(\cos(\theta) + \frac{\sqrt{2}}{2})}{\sqrt{2}}.
  8. Substitute and Simplify: Substitute θ=π4\theta = \frac{\pi}{4} into the simplified function.\newlineNow that we have simplified the function, we can substitute θ=π4\theta = \frac{\pi}{4} again. This gives us 2(cos(π4)+22)2\frac{2(\cos(\frac{\pi}{4}) + \frac{\sqrt{2}}{2})}{\sqrt{2}}. Since cos(π4)=22\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}, the expression simplifies to 2(22+22)2\frac{2(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2})}{\sqrt{2}}.
  9. Final Answer: Simplify the expression.\newlineThe expression simplifies to (2(22/2))/(2)(2(2\sqrt{2}/2))/(\sqrt{2}), which is (22)/(2)(2\sqrt{2})/(\sqrt{2}). The 2\sqrt{2} in the numerator and denominator cancel each other out, leaving us with the final answer of 22.

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