Question

I need to solve this using l'hopital's rule and logarithmic diferentiation.

Answer 1

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.