Question

Limits questions using the lim (sinx/x) = 1 as x approaches 0 theorem.

https://www.dropbox.com/s/1ymbuzawmdz1rsc/20140213_120406.jpg

Answer 1

Can someone show me the first few steps for the first question? I assume the other ones are done in a similar fashion.

Answer 2

the 2nd and the 3rd one are easy, not so sure about the 1 one to be honest.

Answer 3

care to see 2), 3)? For 1) I need some additional time to think it through, seems to be a dead end though with the fundamental limes though.

Answer 4

@insidesin , for 1) use the following:
\[\large \tan (2x)= \tan(2x-\pi) \tag{*} \]
where * is the periodicity of tan. Then you can obtain the following result:
\[\large \lim_{x \to \frac{\pi}{2}} \frac{2 \tan(2x-\pi)}{2x-\pi} = 2 \]

Answer 5

\[\lim_{x \rightarrow \frac{ \pi }{2 }}\frac{ \tan 2x }{x-\frac{ \pi }{ 2 } },put~ x=\frac{ \pi }{ 2 }+h,h \rightarrow~0~as~x \rightarrow \frac{ \pi }{ 2 }\]
\[=\lim_{h \rightarrow 0}\frac{ \tan 2\left( \frac{ \pi }{ 2 }+h \right) }{\frac{ \pi }{2 }+h-\frac{ \pi }{2 } }\]
\[=\lim_{h \rightarrow 0}\frac{ \tan \left( \pi+2h \right) }{ h }=2*\lim_{h \rightarrow 0}\frac{ \tan 2h }{ 2h }\]
=2*1=2

Answer 6

Sure i wouldn't mind seeing the solutions for questions 2 and 3 as well. Do they also use the "hint" that was given? or is it straight trig identities? Pardon me i'm just finishing up different questions on the worksheet before I look at this

Answer 7

No @insidesin, they all work the same way. But I strongly recommend to try them yourself first, in my opinion 1) was the toughest one to master, because of the identity I have mentioned above, or the substitution method provided by @surjithayer.

Problems 2) and 3) aren't hard at all and barely worth to walk through into detail. Use the fundamental limes for 2) as given in your hint, also understand that this hint does not apply for \(\cos x / x \) and for the 3rd one, write \(x^2-4=(x-2)(x+2)\) and you're done.

Answer 8

2.
\[\lim_{x \rightarrow 0}\frac{ \sin 5x \cos 2x }{3x ^{2}+5x }=\frac{ 1 }{2 }\lim_{x \rightarrow 0}\frac{ 2\sin 5x \cos 2x }{3x ^{2}+5x }\]
\[=\frac{ 1 }{ 2 }\lim_{x \rightarrow 0}\frac{ \sin \left( 5x+2x \right)+\sin \left( 5x-2x \right) }{ 3x ^{2}+5x }\]
\[=\frac{ 1 }{ 2 }\left[ \lim_{x \rightarrow 0}\frac{7 \sin 7x }{ 7x \left( 3x+5 \right) }+\lim_{x \rightarrow 0}\frac{ 3\sin 3x }{3 x \left( 3x+5 \right) } \right]\]
\[=\frac{ 7 }{2 }\lim_{x \rightarrow 0}\frac{ \sin 7x }{7x }* \lim_{x \rightarrow0}\frac{ 1 }{3x+5 }+\frac{ 3 }{ 2 }\frac{ \sin 3x }{ 3x }\lim_{x \rightarrow 0}\frac{ 1 }{3x+5 }\]
i think now you can complete.

Answer 9

Also for 2):
\[\large \lim_{x \to 0} \frac{\sin (5x)\cos(2x) 5x}{(5x)(3x^2+5x)}=\lim_{x \to 0} \frac{\sin(5x)}{5x}\cdot \lim_{x \to 0} \cos (2x) \cdot \lim_{x \to 0}\frac{5x}{3x^2+5x} \]
So you're left with 1 times 1 times the limes of \[\large \lim_{x \to 0} \frac{5x}{3x^2+5x}= \lim_{x \to 0} \frac{5}{3x+5} \]

Answer 10

Thanks guys! Both of you were really helpful in assisting me with this question. I wish I could select mutiple correct answers but I'd have to say that I liked surjithayer's substitution method to solving Q1 more. Both of you though were very clear in your answers! Thanks again

Answer 11

\[=\frac{ 7 }{ 2 }*1*\frac{ 1 }{ 0+5 }+\frac{ 3 }{ 2 }*1*\frac{ 1 }{ 0+5 }=\frac{ 7 }{ 10 }+\frac{ 3 }{ 10 }=1\]

Answer 12

yw