How do I rationalize this?

Calculus Level 2

Evaluate $\displaystyle \lim_{x \rightarrow 0} \dfrac{\sin x - \cos x + 1}{\sin x + \cos x -1}$ .

If the answer cannot be inputted into the solution box, select the correct option below and input the number of the option that best fits the correct answer:

Limit DNE, by approaching $+ \infty$ on both sides of zero: Input 1234
Limit DNE, by approaching $- \infty$ on both sides of zero: Input 4321
Limit DNE, by not approaching the same value from left and from right of zero, but not approaching $\pm \infty$ on either side: Input 1324
Limit DNE, by approaching $-\infty$ from the left of zero, and $+ \infty$ from the right of zero: Input 1243
Limit DNE, by approaching $+\infty$ from the left of zero, and $- \infty$ from the right of zero: Input 3241

The answer is 1.

1 solution

Mohammed Elnaggar
Apr 28, 2019

First let's start with conjugate of denominator as treating cos-1 as one term $\frac{(sinx-cosx+1)(sinx-cosx+1)}{(sinx+cosx-1) (sinx-cosx+1) }$ You find that $\frac{(sinx-cosx+1) ^2}{sinx^-(cosx-1) ^2}$ . Then cosx=1,then sinx=0 you will find indeterminate form, and if you tried treat sinx+cosx =1 as one term to make the difference between the two squares or getting the conjugation of numerator as treating cos+1 or sinx - cosx as one term, then the same inderminate form But if you did $\frac{sinx(1+sinx)+cosx(cosx-1)}{sinx(1-sinx)-cosx(cosx-1)}$ by this famous rule sinx +cosx=1 and then group sines and cosines together, then you will multiply numerator and denominator by (1-sinx) which is conjugate of 1+sinx,, and then you will get 1 by cosx=1 and sinx=0 after this multiplication

How do I rationalize this?

The answer is 1.

1 solution

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