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ANTONII [103]
3 years ago
15

I need help with this question x2 - 16y2

Mathematics
1 answer:
lana66690 [7]3 years ago
5 0
A^2 - b^2 = (a+b) (a-b) where a = x and b = 4y
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Answer:

Option a.

\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1

Step-by-step explanation:

You have the following limit:

\lim_{x \to \frac{\pi}{2}{(3e)^{xcosx}

The method of direct substitution consists of substituting the value of \frac{\pi}{2} in the function and simplifying the expression obtained.

We then use this method to solve the limit by doing x=\frac{\pi}{2}

Therefore:

\lim_{x \to \frac{\pi}{2}}{(3e)^{xcosx} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})}

cos(\frac{\pi}{2})=0\\

By definition, any number raised to exponent 0 is equal to 1

So

\lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}(0)}\\\\

\lim_{x\to \frac{\pi}{2}}{(3e)^{0}} = 1

Finally

\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1

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