Question

solve both questions......


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Answer 1

Answer: 1) 1

2) 2

Step-by-step explanation:

1) $\lim_{x \to 0} \frac{sin(ax)+bx}{ax+sin(bx)}$

Dividing numerator and denominator by x, we get

$\lim_{x \to 0} \frac{sin(ax)/x+b}{a+sin(bx)/x}$

(sorry, you'll have to cope up with the clumsy formatting; Brainly doesn't provide a Latex like typing experience as on MSE and Quora, but I hope you'll understand)

For ease in formatting, I'll solve numerator and denominator seperately and merge the two in the end.

Numerator:

$\lim_{x \to 0} \frac{sin(ax)}{ax} a +  \lim_{x\to 0} b$

$=a(1)+b = a+b$ (from the identity lim(sinx)/x as x->0 =1)

Denominator:

$\lim_{x \to 0} a+ \lim_{x \to 0} \frac{sin(bx)}{bx}.b$

$=a+b(1) = a+b$

Now, putting the numerator back over the denominator,

$=\frac{a+b}{a+b}$

=1

2) Let

$y=x-\frac{\pi }{2}$

Then, as x->π/2, y->0. Also, x = π/2 + y

Now,

$\lim_{x \to \frac{\pi }{2} }\frac{tan(2x)}{x-\frac{\pi }{2} }$

$=  \lim_{y \to 0}\frac{tan2(\frac{\pi }{2}+y) }{y}$

$= \lim_{y \to 0} \frac{tan2y}{y} = 2\lim_{y \to 0} \frac{tan2y}{2y}  = 2(1)$

=2