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Masteriza [31]
3 years ago
11

Help I need this now

Mathematics
1 answer:
eimsori [14]3 years ago
3 0

Answer:

There is no question showing up so I dont know how to help u

Step-by-step explanation:

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arlik [135]

Yo sup??

For our convenience let h=x+1

therefore

when x tends to -1, h tends to 0

hence we can rewrite it as

\lim_{h \to \ 0 } (cos(h))^{(cot(h^2 )}

This inequality is of the form 1∞

We will now apply the formula

e^(^g^(^x^)^(^f^(^x^)^-^1^)^)

plugging in the values of g(x) and f(x)

e^{lim_{h \to \ 0}{(cot(h)^2(cos(h)-1))}

express coth² as cosh²/sinh² and also write cosh-1 as 2sin²(h/2)

(by applying the property that cos2x=1-sin²x)

After this multiply the numerator and denominator with h² so that we can apply the property that

\lim_{x \to \ 0 } sinx/x =1

Now your equation will look like this.

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2)*h^2)/(sin(h)^2*h^2)}

We will now apply the result

\lim_{x \to \ 0 } sinx/x =1

where x=h²

we get

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2)}

we now multiply the numerator and denominator with 4 so that we can say

\lim_{h^2 \to \ 0 } sin^2(h/2)/(h^2/4) = 1

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2*4/4)}

=e^{lim_{h \to \ 0}{((cos(h)^2*2)/(4)}

Apply the limits and you will get

e^{cos(0)^2*2/4

=e^{1/2}

Hope this helps.

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Step-by-step explanation:

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Step-by-step explanation:

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