Evaluate: limθ→π4cos⁡θ−sin⁡θθ−π4 My Attempt: This limit takes 00 form when θ=π4. Here, 00 form is an indeterminate form. So how do I make it determinate?

Since your limit is in the form 00, you can use LHopitals rule. limθ→π4cos⁡θ−sin⁡θθ−π4=limθ→π4ddθ(cos⁡θ−sin⁡θ)ddθ(θ−π4)=limθ→π4(−sin⁡θ−cos⁡θ)=… Now that your limit is not indeterminate, you can evaluate your limit by substitution.

If you know that for small x you have sin⁡x≈x (from Taylor expansion or l'Hopital), you can write θ−π4=x. Then you can show cos⁡(x+π4)−sin⁡(x+π4)=−2sin⁡x Then your limit is −2

Since your limit is in the form 00, you can use LHopitals rule. limθ→π4cos⁡θ−sin⁡θθ−π4=limθ→π4ddθ(cos⁡θ−sin⁡θ)ddθ(θ−π4)=limθ→π4(−sin⁡θ−cos⁡θ)=… Now that your limit is not indeterminate, you can evaluate your limit by substitution.

If you know that for small x you have sin⁡x≈x (from Taylor expansion or l'Hopital), you can write θ−π4=x. Then you can show cos⁡(x+π4)−sin⁡(x+π4)=−2sin⁡x Then your limit is −2

Step by step solution to "Evaluate: limθ→π4cos⁡θ−sin⁡θθ−π4 My Attempt: This limit takes 00..."

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deiteresfp

Answered question

2021-12-28

Evaluate: $lim_{θ \to \frac{π}{4}} \frac{\cos θ - \sin θ}{θ - \frac{π}{4}}$
My Attempt:
This limit takes $\frac{0}{0}$ form when $θ = \frac{π}{4}$ . Here, $\frac{0}{0}$ form is an indeterminate form. So how do I make it determinate?

Answer & Explanation

nghodlokl

Beginner2021-12-29Added 33 answers

Without lHospital:

vicki331g8

Beginner2021-12-30Added 37 answers

Since your limit is in the form

\frac{00}{}

, you can use LHopitals rule.

lim_{θ \to \frac{π}{4}} \frac{\cos θ - \sin θ}{θ - \frac{π}{4}} = lim_{θ \to \frac{π}{4}} \frac{\frac{d}{d θ} (\cos θ - \sin θ)}{\frac{d}{d θ} (θ - \frac{π}{4})} = lim_{θ \to \frac{π}{4}} (- \sin θ - \cos θ) = \dots

Now that your limit is not indeterminate, you can evaluate your limit by substitution.

nick1337

Expert2022-01-08Added 777 answers

If you know that for small x you have $\sin x \approx x$ (from Taylor expansion or l'Hopital), you can write $θ - \frac{π}{4} = x$ . Then you can show
$\cos (x + \frac{π}{4}) - \sin (x + \frac{π}{4}) = - \sqrt{2} \sin x$
Then your limit is $- \sqrt{2}$

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