MHB How Can You Solve This Limit Problem Without L'Hospital's Rule?

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To solve the limit problem $$ \lim_{x\to -2} \sin(\frac{\pi x}{2})\frac{x^2+1}{x+2} $$ without L'Hospital's rule, one can use the limit property $$\lim_{u \to 0} \frac{\sin{u}}{u} = 1.$$ By applying partial fraction decomposition to the rational function and expanding the limit, two of the three terms can be evaluated directly. The remaining term can be manipulated into the form of the known limit by substituting appropriately. This approach provides a method to determine the limit without resorting to L'Hospital's rule.
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I would really appreciate if you could help me solving this limit problem!

Determine the limit without using L'Hospital's rule!

$$ \lim_{x\to -2} \sin(\frac{\pi x}{2})\frac{x^2+1}{x+2} = ?$$

Thank you in advance!
 
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Do you know or have proved (or in general are allowed to use) the limit below?
$$\lim_{u \to 0} \frac{\sin{u}}{u} = 1$$
If so, apply partial fraction decomposition to the rational function (the quotient of the two polynomials; that's not actually a polynomial) and expand the limit. You can then evaluate two of the three terms, and for the remaining term try and get it in the form of the limit above for some $u$ to determine the limit of your expression. Good luck!
 
zadir said:
I would really appreciate if you could help me solving this limit problem!

Determine the limit without using L'Hospital's rule!

$$ \lim_{x\to -2} \sin(\frac{\pi x}{2})\frac{x^2+1}{x+2} = ?$$

Thank you in advance!

Take into account the simple identity...

$\displaystyle \sin \frac{\pi\ x}{2} = - \sin (\frac{\pi\ x}{2} + \pi) = - \sin [\frac{\pi}{2}\ (x+ 2)]\ (1)$

Kind regards

$\chi$ $\sigma$
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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