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Find 
lim_(x rarr(pi)/(4))(cos(2x))/(cos(x)-sin(x))
Choose 1 answer:
(A) 
sqrt2
(B) 2
(c) 4
(D) The limit doesn't exist

Find limxπ4cos(2x)cos(x)sin(x) \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos (2 x)}{\cos (x)-\sin (x)} .\newlineChoose 11 answer:\newline(A) 2 \sqrt{2} \newline(B) 22\newline(C) 44\newline(D) The limit doesn't exist

Full solution

Q. Find limxπ4cos(2x)cos(x)sin(x) \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos (2 x)}{\cos (x)-\sin (x)} .\newlineChoose 11 answer:\newline(A) 2 \sqrt{2} \newline(B) 22\newline(C) 44\newline(D) The limit doesn't exist
  1. Substitute Values: We need to find the limit of the function (cos(2x))/(cos(x)sin(x))(\cos(2x))/(\cos(x)-\sin(x)) as xx approaches π/4\pi/4. Let's first try to directly substitute x=π/4x = \pi/4 into the function to see if it yields a determinate form.\newlinecos(2×(π/4))=cos(π/2)=0\cos(2 \times (\pi/4)) = \cos(\pi/2) = 0\newlinecos(π/4)=2/2\cos(\pi/4) = \sqrt{2}/2\newlinesin(π/4)=2/2\sin(\pi/4) = \sqrt{2}/2\newlineNow, substitute these values into the function:\newline(cos(2×(π/4)))/(cos(π/4)sin(π/4))=0/(2/22/2)=0/0(\cos(2 \times (\pi/4)))/(\cos(\pi/4) - \sin(\pi/4)) = 0/(\sqrt{2}/2 - \sqrt{2}/2) = 0/0\newlineWe get an indeterminate form 0/00/0, which means we need to apply L'Hôpital's Rule or algebraic manipulation to find the limit.
  2. Apply L'Hôpital's Rule: Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit of f(x)/g(x)f(x)/g(x) as xx approaches a value cc is 0/00/0 or /\infty/\infty, then the limit of f(x)/g(x)f(x)/g(x) as xx approaches cc is the same as the limit of f(x)/g(x)f'(x)/g'(x) as xx approaches cc, provided that the latter limit exists.\newlineLet's find the derivatives of the numerator and the denominator.\newlineThe derivative of xx11 with respect to xx is xx33.\newlineThe derivative of xx44 with respect to xx is xx66.
  3. Find Derivatives: Now we will apply L'Hôpital's Rule by taking the limit of the derivatives: \newlinelimxπ4(2sin(2x)sin(x)cos(x))\lim_{x \to \frac{\pi}{4}}\left(\frac{-2\sin(2x)}{-\sin(x) - \cos(x)}\right)\newlineLet's substitute x=π4x = \frac{\pi}{4} into the derivatives:\newline2sin(2×(π4))=2sin(π2)=2-2\sin(2 \times (\frac{\pi}{4})) = -2\sin(\frac{\pi}{2}) = -2\newlinesin(π4)cos(π4)=2222=2-\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}\newlineNow, we substitute these values into the limit of the derivatives:\newline22=22=2×(22)=2\frac{-2}{-\sqrt{2}} = \frac{2}{\sqrt{2}} = 2 \times (\frac{\sqrt{2}}{2}) = \sqrt{2}
  4. Apply L'Hôpital's Rule Again: We have found that the limit of the original function as xx approaches π4\frac{\pi}{4} is 2\sqrt{2}. Therefore, the correct answer is (A) 2\sqrt{2}.

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