Competition - post a photo you've taken of nature >>

FP3 - 2 Limits Questions - Stuck

Ready To Answer Now - Latex Problem Sorted
If I didn't rep you that's because I couldn't

Q1:

\displaystyle\lim_{x\to \infty} \dfrac{x+e^x}{x-e^x}

I have tried multiplying top and bottom by

x + e^x

and that didn;t get me anywhere, I then tried expanding the e^x's and after some working eventually got stuck at

Unparseable latex formula:

\displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}[br]\[br]= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}

I have been on it for a while so a solution would be better than a hint, unless there is something small that I am missing or a mistake somewhere.

Q2:

Unparseable latex formula:

\displaystyle F(x) = \dfrac{e^x}{1-x}[br]\[br]\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty

For this one I expanded e^x and brought the (1-x) to the top and expanded it, multiplied it out and I am stuck at

(1+2x+\frac{5x^2}{2} \cdots)

Same as above, all help is appreciated.

Thnx

(edited 11 years ago)

Reply 1

ian.slater

On Q1 you could try multiplying top and bottom by e^-x.

Reply 2

member910132

Original post by ian.slater

On Q1 you could try multiplying top and bottom by e^-x.

Yea, that works lol. The answer is -1, why didn't I see something so simple !!

Reply 3

Dirac Spinor

Original post by member910132

Ready To Answer Now - Latex Problem Sorted

Q1:

\displaystyle\lim_{x\to \infty} \dfrac{x+e^x}{x-e^x}

I have tried multiplying top and bottom by

x + e^x

and that didn;t get me anywhere, I then tried expanding the e^x's and after some working eventually got stuck at

Unparseable latex formula:

\displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}[br]\[br]= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}

I have been on it for a while so a solution would be better than a hint, unless there is something small that I am missing or a mistake somewhere.

Q2:

Unparseable latex formula:

\displaystyle F(x) = \dfrac{e^x}{1-x}[br]\[br]\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty

For this one I expanded e^x and brought the (1-x) to the top and expanded it, multiplied it out and I am stuck at

(1+2x+\frac{5x^2}{2} \cdots)

Same as above, all help is appreciated.

Thnx

second one:
large +ve x makes e^x +ve and 1-x -ve so F is -ve. Now consider the relative order of exponentials and linear expressions in x.

Reply 4

member910132

Original post by ian.slater

On Q1 you could try multiplying top and bottom by e^-x.

Just tried that for Q2 and I get:

\displaystyle\dfrac{1}{e^{-x}-xe^{-x}}

Using the general result that xe^(-x) tends to 0 as x tends to infinity my only problem is showing how F(x) goes to negative infinity and not just infinity.

(edited 11 years ago)

Reply 5

mikelbird

Original post by member910132

Ready To Answer Now - Latex Problem Sorted

Q1:

\displaystyle\lim_{x\to \infty} \dfrac{x+e^x}{x-e^x}

I have tried multiplying top and bottom by

x + e^x

and that didn;t get me anywhere, I then tried expanding the e^x's and after some working eventually got stuck at

Unparseable latex formula:

\displaystyle \dfrac{1+2x+\dfrac{x^2}{2}+...}{-1-\frac{x^2}{2}+\cdots}[br]\[br]= (-1)(1+2x+\dfrac{x^2}{2} + \cdots)(1+ (\dfrac{x^2}{2} + \dfrac{x^3}{6}+\cdots))^{-1}

I have been on it for a while so a solution would be better than a hint, unless there is something small that I am missing or a mistake somewhere.

Q2:

Unparseable latex formula:

\displaystyle F(x) = \dfrac{e^x}{1-x}[br]\[br]\text{Show that} F(x) \to -\infty \ \text {as} \ x \to \infty

For this one I expanded e^x and brought the (1-x) to the top and expanded it, multiplied it out and I am stuck at

(1+2x+\frac{5x^2}{2} \cdots)

Same as above, all help is appreciated.

Thnx

You could use L'Hopital' s Theorem...they both come out very easily then.....

Reply 6

member910132

Original post by ben-smith

second one:
large +ve x makes e^x +ve and 1-x -ve so F is -ve. Now consider the relative order of exponentials and linear expressions in x.

Sorry, I don't follow the second part.

Reply 7

member910132

Original post by mikelbird

You could use L'Hopital' s Theorem...they both come out very easily then.....

That isn't on the AQA FP3 syllabus and I don't want to learn stuff that ain't on the syllabus.

Reply 8

Dirac Spinor

Original post by member910132

Sorry, I don't follow the second part.

from an intuitive point of view, e^x is like a polynomial of very high degree and so it's going to tend to infinity faster than a linear function.

Reply 9

ian.slater

Original post by member910132

Just tried that for Q2 and I get:

\displaystyle\dfrac{1}{e^{-x}-xe^{-x}}

Using the general result that xe^(-x) tends to 0 as x tends to infinity my only problem is showing how F(x) goes to negative infinity and not just infinity.