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?
$$L_1=\lim_{x \to 0} \frac {(1+2x)^{1/3}-1}{x}$$


Using binomail thorem: $$(1+2x)^{1/3}=1+\frac13.2x+\text{terms containing other higher powers of x}$$
Putting back in you'll get the limit as $2/3$




  $$L_2=\lim_{x \to 5^-} \frac {e^x}{(x-5)^3}=-\infty$$


Because denominator $\to0^-$ and the numerator is finite.




  $$L_3=\lim_{x \to \pi/2} \frac{\sin(x)}{\cos(x)^2} - \tan(x)^2$$


Since,
$$\frac{\sin(x)}{\cos(x)^2} - \tan(x)^2=\frac{\sin x-\sin^2 x}{\cos^2x}=\frac{\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\frac{\sin x}{1+\sin x}$$
So, obviously limit is $\frac 12$
EXPAND(1+2x)^3/2 and the limit will reduce to 2/3  HOWZ THAT