<s>[INST] Limits without L'Hopitals Rule Evaluate the limit without using L'hopital's rule
a)$$\lim_{x \to 0} \frac {(1+2x)^{1/3}-1}{x} $$
I got the answer as $l=\frac 23$... but I used L'hopitals rule for that... How can I do it another way?
b)$$\lim_{x \to 5^-} \frac {e^x}{(x-5)^3}$$ 
$l=-\infty$
c)$$\lim_{x \to \frac {\pi} 2} \frac{\sin x}{\cos^2x} - \tan^2 x$$ 
I don't know how to work with this at all
So basically I was able to find most of the limits through L'Hopitals Rule... BUT how do I find the limits without using his rule?
 [/INST] a) $$\lim_{x\to0}{\frac{\big(1+2x\big)^{\frac{1}{3}}-1}{x}}$$
First, let $f(x)=\big(1+2x\big)^{\frac{1}{3}}$, we know that $f(0)=1$ and $f'(x)=\frac{2}{3\big(1+2x\big)^{\frac{2}{3}}}$ $\Rightarrow$ $f'(0)=\frac{2}{3}$
so $$\lim_{x\to0}{\frac{\big(1+2x\big)^{\frac{1}{3}}-1}{x}}$$
$$=\lim_{x\to0}{\frac{f(x)-f(0)}{x-0}}=f'(0)=\frac{2}{3}$$
b) $$\lim_{x\to5^-}{\frac{e^x}{\big(x-5\big)^3}}$$
$$=\lim_{x\to5^-}{e^x}\cdot\lim_{x\to5^-}{\frac{1}{\big(x-5\big)^3}}$$
$$=e^5\cdot-\infty=-\infty$$
c) $$\lim_{x\to\frac{\pi}{2}}{\frac{\sin x}{\cos^2 x}-\tan^2 x}=L$$
Let's multiply by the conjugate of $\frac{\sin x}{\cos^2 x}-\tan^2 x$:
$$L=\lim_{x\to\frac{\pi}{2}}{\frac{\big(\frac{\sin x}{\cos^2 x}-\tan^2 x\big)\big(\frac{\sin x}{\cos^2 x}+\tan^2 x\big)}{\frac{\sin x}{\cos^2 x}+\tan^2 x}}$$
By smplifying the numerator we get:
$\big(\frac{\sin x}{\cos^2 x}-\tan^2 x\big)\big(\frac{\sin x}{\cos^2 x}+\tan^2 x\big)=\frac{\sin^2 x}{\cos^4 x}-\tan^4 x$
$=\tan^2 x\big(\frac{1-\sin^2 x}{\cos^2}\big)=\tan^2 x$
$$\Rightarrow L=\lim_{x\to\frac{\pi}{2}}{\frac{\tan^2 x}{\frac{\sin x}{\cos^2 x}+\tan^2 x}}$$
$$=\lim_{x\to\frac{\pi}{2}}{\frac{\sin x}{1+\sin x}}=\frac{1}{2}$$
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