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In calculus, while solving for a limit we do some manipulations, Consider an example,

$$\lim_{x\to 0}\frac{(1+2x)^{0.5}-1}{x}$$

$$=\lim_{x\to 0}\frac{2}{(1+2x)^{0.5}+1}$$

Then we put $x = 0$ and solve. But, didn't this manipulation change the domain of the function? As we can see that in the original question $x$ was $\not=0$ but in second case $x=0$ is possible.

So does that mean that it is not important for the functions to be identical to find its limit? And if yes then how can we be sure that the limit we have calculated is correct?

Harish Chandra Rajpoot
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Preet
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  • I think you should look up "functions that agree at all but at one point" on Google. Both arguments of the limit expressions are equal for all $x>0$, but the second argument is defined at $x=0$ while the first one is not. This makes it possible to evaluate the limit of the first one using the exact value of the second. – YuJin Kim Jul 30 '20 at 05:12

3 Answers3

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In this case I'm assuming you know how to show that the limit indeed exists.

Let $L_1 = \lim_{x\to 0} \frac{\sqrt{1+2x}-1}{x} = \lim_{x \to 0} f(x)$

Let $L_2 = \lim_{x \to 0}\frac{\sqrt{1+2x} + 1}{\sqrt{1+2x} + 1} = \lim_{x \to 0} g(x) = 1$

Now, we use the property that if both $L_1, L_2$ exist, then

$$\lim_{x\to 0} f(x)g(x) = L_1L_2 = L_1$$

Hence though the two functions are not exactly identical, their limits are equal

Dhanvi Sreenivasan
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The key difference in the above two cases is whether the function is continuous at the point where the limit is evaluated, namely consider $$ f(x) = \frac{(1+2x)^{0.5}-1}{x} \qquad \text{and} \qquad g(x) = \frac{2}{(1+2x)^{0.5}+1}.$$ Then $g$ is continuous at 0. By definition of continuity, we have $$\lim\limits_{x \to 0} g(x) = g(0).$$ i.e. you can directly substitute 0 to evaluate the limit. However, $f$ is not continuous at 0, so you cannot perform such operation to evaluate the limit and must use other techniques.

Remark: Finding a limit does not require the function to be defined at that point. What is required is the definition of the function 'near' the point.

Ken Hung
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The manipulation

$$\frac{\sqrt{1+2x}-1}{x}\cdot \frac{\sqrt{1+2x}+1}{\sqrt{1+2x}+1}=\frac 2{\sqrt{1+2x}+1}$$

doesn't change the function because we are multipling it by $1$, $\forall x>-\frac12$ and therefore that manipulation doesn't affect the value for the limit at $x=0$.

See also the related:

user
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