Answer :
To evaluate the limit [tex]\(\lim_{x \rightarrow 1} \frac{\sqrt{1+x^2}-\sqrt{2}}{x-1}\)[/tex], we begin by checking if there's an immediate substitution of [tex]\(x=1\)[/tex] that results in an indeterminate form.
1. Substitute [tex]\(x = 1\)[/tex] into the function:
[tex]\[ \frac{\sqrt{1+1^2}-\sqrt{2}}{1-1} = \frac{\sqrt{2}-\sqrt{2}}{0} = \frac{0}{0} \][/tex]
Since it results in the indeterminate form [tex]\(\frac{0}{0}\)[/tex], we need to manipulate the expression to evaluate the limit. A common technique for such problems is to multiply by the conjugate of the numerator.
2. Multiply the numerator and the denominator by the conjugate of [tex]\(\sqrt{1+x^2}-\sqrt{2}\)[/tex]:
[tex]\[ \frac{\sqrt{1+x^2}-\sqrt{2}}{x-1} \times \frac{\sqrt{1+x^2}+\sqrt{2}}{\sqrt{1+x^2}+\sqrt{2}} \][/tex]
This gives:
[tex]\[ \frac{(\sqrt{1+x^2}-\sqrt{2})(\sqrt{1+x^2}+\sqrt{2})}{(x-1)(\sqrt{1+x^2}+\sqrt{2})} \][/tex]
3. Simplify the numerator using the difference of squares:
[tex]\[ (\sqrt{1+x^2})^2 - (\sqrt{2})^2 = (1+x^2) - 2 = x^2 - 1 \][/tex]
Thus, the expression becomes:
[tex]\[ \frac{x^2 - 1}{(x-1)(\sqrt{1+x^2}+\sqrt{2})} \][/tex]
4. Factor the numerator [tex]\(x^2 - 1\)[/tex]:
[tex]\[ x^2 - 1 = (x-1)(x+1) \][/tex]
Now the expression looks like:
[tex]\[ \frac{(x-1)(x+1)}{(x-1)(\sqrt{1+x^2}+\sqrt{2})} \][/tex]
5. Cancel out the common term [tex]\((x-1)\)[/tex]:
[tex]\[ \frac{x+1}{\sqrt{1+x^2}+\sqrt{2}} \][/tex]
6. Now, substitute [tex]\(x = 1\)[/tex] into the simplified expression:
[tex]\[ \frac{1+1}{\sqrt{1+1^2}+\sqrt{2}} = \frac{2}{\sqrt{2}+\sqrt{2}} = \frac{2}{2\sqrt{2}} \][/tex]
7. Simplify the fraction:
[tex]\[ \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \][/tex]
8. Rationalize the denominator:
[tex]\[ \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \rightarrow 1} \frac{\sqrt{1+x^2}-\sqrt{2}}{x-1} = \frac{\sqrt{2}}{2} \][/tex]
1. Substitute [tex]\(x = 1\)[/tex] into the function:
[tex]\[ \frac{\sqrt{1+1^2}-\sqrt{2}}{1-1} = \frac{\sqrt{2}-\sqrt{2}}{0} = \frac{0}{0} \][/tex]
Since it results in the indeterminate form [tex]\(\frac{0}{0}\)[/tex], we need to manipulate the expression to evaluate the limit. A common technique for such problems is to multiply by the conjugate of the numerator.
2. Multiply the numerator and the denominator by the conjugate of [tex]\(\sqrt{1+x^2}-\sqrt{2}\)[/tex]:
[tex]\[ \frac{\sqrt{1+x^2}-\sqrt{2}}{x-1} \times \frac{\sqrt{1+x^2}+\sqrt{2}}{\sqrt{1+x^2}+\sqrt{2}} \][/tex]
This gives:
[tex]\[ \frac{(\sqrt{1+x^2}-\sqrt{2})(\sqrt{1+x^2}+\sqrt{2})}{(x-1)(\sqrt{1+x^2}+\sqrt{2})} \][/tex]
3. Simplify the numerator using the difference of squares:
[tex]\[ (\sqrt{1+x^2})^2 - (\sqrt{2})^2 = (1+x^2) - 2 = x^2 - 1 \][/tex]
Thus, the expression becomes:
[tex]\[ \frac{x^2 - 1}{(x-1)(\sqrt{1+x^2}+\sqrt{2})} \][/tex]
4. Factor the numerator [tex]\(x^2 - 1\)[/tex]:
[tex]\[ x^2 - 1 = (x-1)(x+1) \][/tex]
Now the expression looks like:
[tex]\[ \frac{(x-1)(x+1)}{(x-1)(\sqrt{1+x^2}+\sqrt{2})} \][/tex]
5. Cancel out the common term [tex]\((x-1)\)[/tex]:
[tex]\[ \frac{x+1}{\sqrt{1+x^2}+\sqrt{2}} \][/tex]
6. Now, substitute [tex]\(x = 1\)[/tex] into the simplified expression:
[tex]\[ \frac{1+1}{\sqrt{1+1^2}+\sqrt{2}} = \frac{2}{\sqrt{2}+\sqrt{2}} = \frac{2}{2\sqrt{2}} \][/tex]
7. Simplify the fraction:
[tex]\[ \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \][/tex]
8. Rationalize the denominator:
[tex]\[ \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \rightarrow 1} \frac{\sqrt{1+x^2}-\sqrt{2}}{x-1} = \frac{\sqrt{2}}{2} \][/tex]