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ApoorvGoel · Apr 25, 2011, 02:15 PM

lim(x-->1/0) [(x^2+5x+3)/(x^2+x+2)]^x is equal to

1.) e^4
2.) e^2
3.) e^3
4.) e

please give your answer with correct explanations...

ebaines · Apr 25, 2011, 02:40 PM

This is what you asked:

$ \displaystyle \lim _{x \to 1/0} ( \frac {x^2 + 5x+3} {x^2 + x+2} )^x $

What does x -->1/0 mean? It's pretty easy to show that the value of this for x =0 is 1, and the value for x = 1 is 2.25, so I suspect you meant to write something else.

ApoorvGoel · Apr 25, 2011, 02:53 PM

1/0 means infinite

ApoorvGoel · Apr 25, 2011, 02:55 PM

Now please give your answer with explanation

ebaines · Apr 25, 2011, 03:10 PM

I'll let you figure the full explanation, but here's a hint:

If you divide (x^2 + 5x + 3) by (x^2 + x + 2) you get:

$ 1 + \frac {4x + 1}{x^2 + x + 2} = 1 + \frac 4 {x + 1 + 2/x} + \frac 1 {x^2 + x + 2}$

What does this become for large values of x?

Now recall that

$ \lim _{n \to \infty} ( 1 + \frac i n )^n = e^n $

Can you take it from here?

ApoorvGoel · Apr 26, 2011, 03:22 AM

Sir I had already up to yhis point but after that I am not able to attempt anything

ebaines · Apr 26, 2011, 05:49 AM

Consider what happens to

$ 1 + \frac 4 {x + 1 + 2/x} + \frac 1 {x^2 + x + 2} $
as x gets very large. The term $x^2$ goes to infinity much faster than $x$ , and $\frac 1 x$ goes to zero.

Therefore $1 + \frac 4 {x + 1 + 2/x} + \frac 1 {x^2 + x + 2}$ approaches $1 + \frac 4 x$ for large values of $x$ .

So now you have

$ \lim _{x \to \infty} (1 + \frac 4 {x + 1 + 2/x} + \frac 1 {x^2 + x + 2}) = \lim _{x \to \infty} (1 + \frac 4 x ) $

The rest is straight forward, using the formula for $e^n$ that I gave you earlier.

Unknown008 · Apr 26, 2011, 08:32 AM

sir I had already up to yhis point but after that I am not able to attempt anything

That's why we are asking you to post your attempts so far. It might be difficult to type (but I don't see how more difficult that can be to post the question itself), but in the end, you're saving time for yourself and for us, to avoid the steps you already went through.

ApoorvGoel · Apr 26, 2011, 01:35 PM

Thanks a lot for your answer