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Use continuity to evaluate the limit.

$ \displaystyle \lim_{x \to 1} \ln\biggl( \dfrac{5 - x^2}{1 + x} \biggr) $

The function $f(x)=\ln \left(\frac{5-x^{2}}{1+x}\right)$ is continuous throughout its domain because it is the composite of a logarithm function and a rational function. For the domain of $f,$ we must have $\frac{5-x^{2}}{1+x}>0,$ so the numerator and denominator must have the same sign, that is, the domain is $(-\infty,-\sqrt{5}] \cup(-1, \sqrt{5}]$. The number 1 is in that domain, so $f$ is continuous at 1 , and$\lim _{x \rightarrow 1} f(x)=f(1)=\ln \frac{5-1}{1+1}=\ln 2$

02:24

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Campbell University

University of Nottingham

Boston College

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Okay, we want to find the limit of this expression. This function As extra approaches one using continuity. So let's look at the point where dysfunction is continuous. Let's let's look at some restrictions. Now, if we look at five minus x squared over one plus x. Five minus x squared will be a continuous function. One plus X will be continuous function. So a continuous function over a continuous function will be continuous as long as the denominator. Once you start making fractions with continuous functions, you have to make sure that the function in the denominator is not zero. So even though one plus X is a continuous function for all values of X, because one plus X isn't the denominator of this fraction, we have to make sure that one plus X is not zero. So one plus X cannot equal zero. In other words X cannot equal negative one. So X is not allowed to equal negative one because if X was negative one, you would have a zero in the denominator. You cannot divide by zero. Now, another restriction, uh you can only take logarithms of positive numbers. You cannot take the law algorithm of zero. You cannot take the law algorithm of a negative number. So, if we're taking the law algorithm of this expression, this expression must be positive. So five minus X squared. This is just things to keep in mind as we evaluate the limit as X approaches one. Uh this expression has to be positive because we can only take uh natural log of positive numbers. So five minus x squared over one plus X Must be positive, must be greater than zero. So this logarithms um we know that five minus x squared over one plus X is a continuous function when x doesn't equal negative one. For all, other values of x. Five minus x squared over one plus x is continuous. The natural algorithm is a continuous continuous function, composite functions, a function, a continuous function of a continuous function will be continuous as long as uh well not to confuse you, but the natural log of this expression will be continuous as long as this expression is continuous. We know that will be the case as long as X doesn't equal negative one, but the natural log of this expression will be continuous. In other words, this entire function is continuous as long as uh this expression and a parentheses is greater than zero cost law algorithm, which only take a log of positive numbers. So this entire function will be continuous as long as X isn't -1. And as long as uh this little expression in the parenthesis stays positive. So assuming this is a continuous function and we'll find out as X is approaching one how this all behaves. Uh First of all effects is approaching one. Uh That means we're going to have x values near one, so we're not going to have uh an X value equal to negative one effects is getting close to positive one. So we don't have to worry about this condition over here. We do want to pay attention to what's going on in here. This entire function is continuous. Uh As long as this will be positive. Now, if I plug in one into uh this expression here, What does this quotient approach as x approaches one while five minus x squared will approach five minus one squared or five minus one, which is four over and then one plus X as X approaches 11 plus X will approach one plus one or two. so uh this little expression in the princes really approaches 4/2, which of course is to so remember this entire function is continuous. As long as X doesn't equal negative one, we don't have to worry about that because X is going to be close to positive one. And as long as this expression uh is positive when, when x is getting close to one, we see that this expression in parentheses is getting close to two, uh which of course is positive, so for X is close to one. Uh this expression is going to be positive. So this entire function will be a continuous function because X is not going to be negative one. And the low expression and princey sustained positive. So since this is a continuous function we can find a limit Of a continuous function as X approaches one simply by evaluating uh dysfunction at one. And so the limit of the log of five minus X squared over one plus X. As X approaches, one is going to be exactly equal to the log of five minus one squared Over one Plus 1. Five minus one squared over one plus one is 4/2, which is two. So the limit of this function, as X approaches one is the natural log of two.

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