Dividing through by the 4th power of the cosine function, multiplying by 30 and using a Pythagorean identity, you may write:
15\tan^4(x)+10=6\left(\tan^2(x)+1\right)^2
Arrange this as a quadratic in the square of the tangent function...
Let's assume we are taking the limit from the right. Doing the division I suggested, the expression becomes:
\frac{5+3x^2e^{-\frac{5}{x}}}{2\left(1-3xe^{-\frac{5}{x}}\right)}
I do agree that the problem should have stated from the right. I simplified what you have to:
\ln(L)=\lim_{x\to0}\left(\frac{5e^{\frac{5}{x}}+3x^2}{2\left(e^{\frac{5}{x}}-3x\right)}\right)
Now, divide numerator and denominator by:
e^{\frac{5}{x}}...
This is what I had in mind. Begin with the lhs:
\frac{\cos(2x)}{3-\sin(2x)-4\sin^2(x)}
Apply double-angle identities:
\frac{\cos^2(x)-\sin^2(x)}{3-2\sin(x)\cos(x)-4\sin^2(x)}
Divide numerator and denominator by \cos^2(x)
\frac{1-\tan^2(x)}{3\sec^2(x)-2\tan(x)-4\tan^2(x)}
In the...
Try using double angle identities, then divide numerator and denominator by the square of the cosine function , apply a Pythagorean identity in the denominator to get rid of the secant function, then factor and cancel.
The nth triangular number is:
T_n=\sum_{k=1}^{n}(k)
Perhaps the simplest way to find this closed form is to write:
T_n=1+2+3+\dots+(n-2)+(n-1)+n
T_n=n+(n-1)+(n-2)\dots+3+2+1
Adding, we find we have n summands of n + 1:
2T_n=n(n+1)
Thus:
T_n=\frac{n(n+1)}{2}
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