We have been asked to evaluate the following limits:
[tex]$\longrightarrow \lim \limits_{x \to 0} \dfrac{\sin 7x - \sin 4x}{\sin 5x - \sin 3x}$[/tex]
By directly substituting the limit x → 0, we get the following results:
[tex]\dfrac{\sin 7(0) - \sin 4(0)}{\sin 5(0) - \sin 3(0)} = \dfrac{\sin 0 - \sin 0}{\sin 0 - \sin 0} = \dfrac{0 - 0}{0 - 0} = \dfrac{0}{0}\\[/tex]
Since 0/0 is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
[tex]\implies \lim \limits_{x \to 0} \dfrac{\frac{d}{dx}\left(\sin 7x - \sin 4x\right)}{\frac{d}{dx}\left(\sin 5x - \sin 3x\right)}[/tex]
[tex]\implies \lim \limits_{x \to 0} \dfrac{7\cos 7x - 4\cos 4x}{5\cos 5x - 3\cos 3x}[/tex]
Now, evaluate the limits by plugging in 0 for all occurrences of [tex]x[/tex].
[tex]\implies \dfrac{7\cos (7 \cdot 0) - 4\cos (4\cdot 0)}{5\cos (5 \cdot 0) - 3\cos (3\cdot 0)}[/tex]
[tex]\implies \dfrac{7\cos (7 \cdot 0) - 4\cos (4\cdot 0)}{5\cos (5 \cdot 0) - 3\cos (3\cdot 0)}[/tex]i
[tex]\implies \dfrac{7 - 4}{5 - 3}[/tex]
[tex]\implies \dfrac{3}{2}[/tex]
Hence we arrive at a result:
[tex]$\boxed{\lim \limits_{x \to 0} \dfrac{\sin 7x - \sin 4x}{\sin 5x - \sin 3x} = \dfrac{3}{2}}$[/tex]
Answer:
Refer the above attachment
Step-by-step explanation:
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Limits - Calculus
We have been asked to evaluate the following limits:
[tex]$\longrightarrow \lim \limits_{x \to 0} \dfrac{\sin 7x - \sin 4x}{\sin 5x - \sin 3x}$[/tex]
By directly substituting the limit x → 0, we get the following results:
[tex]\dfrac{\sin 7(0) - \sin 4(0)}{\sin 5(0) - \sin 3(0)} = \dfrac{\sin 0 - \sin 0}{\sin 0 - \sin 0} = \dfrac{0 - 0}{0 - 0} = \dfrac{0}{0}\\[/tex]
Since 0/0 is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
[tex]\implies \lim \limits_{x \to 0} \dfrac{\frac{d}{dx}\left(\sin 7x - \sin 4x\right)}{\frac{d}{dx}\left(\sin 5x - \sin 3x\right)}[/tex]
[tex]\implies \lim \limits_{x \to 0} \dfrac{7\cos 7x - 4\cos 4x}{5\cos 5x - 3\cos 3x}[/tex]
Now, evaluate the limits by plugging in 0 for all occurrences of [tex]x[/tex].
[tex]\implies \dfrac{7\cos (7 \cdot 0) - 4\cos (4\cdot 0)}{5\cos (5 \cdot 0) - 3\cos (3\cdot 0)}[/tex]
[tex]\implies \dfrac{7\cos (7 \cdot 0) - 4\cos (4\cdot 0)}{5\cos (5 \cdot 0) - 3\cos (3\cdot 0)}[/tex]i
[tex]\implies \dfrac{7 - 4}{5 - 3}[/tex]
[tex]\implies \dfrac{3}{2}[/tex]
Hence we arrive at a result:
[tex]$\boxed{\lim \limits_{x \to 0} \dfrac{\sin 7x - \sin 4x}{\sin 5x - \sin 3x} = \dfrac{3}{2}}$[/tex]
Answer:
Refer the above attachment
Step-by-step explanation:
Sorry you're answer is deleted
if you are free then answer me
Btw how was your physics exam
Bye...