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lim_((x,y)rarr(0,0))(x^(2)+y^(2))/(sqrt(x^(2)+y^(2)+1)-1)

lim(x,y)(0,0)x2+y2x2+y2+11 \lim _{(x, y) \rightarrow(0,0)} \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1}

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Q. lim(x,y)(0,0)x2+y2x2+y2+11 \lim _{(x, y) \rightarrow(0,0)} \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1}
  1. Substitute r2r^2: Step 11: Simplify the expression by substituting r2r^2 for x2+y2x^2 + y^2, where rr is the distance from the origin in polar coordinates.\newline- Calculation: Let r2=x2+y2r^2 = x^2 + y^2.\newline- Reasoning: This substitution simplifies the expression to a function of a single variable rr.\newline-
  2. Rewrite using rr: Step 22: Rewrite the limit using rr.\newline- Calculation: limr0r2r2+11\lim_{r\to 0} \frac{r^2}{\sqrt{r^2 + 1} - 1}.\newline- Reasoning: This step converts the two-variable limit into a one-variable limit, making it easier to evaluate.\newline-
  3. Apply L'Hopital's Rule: Step 33: Apply L'Hopital's Rule since direct substitution gives 0/00/0, an indeterminate form.\newline- Calculation: Differentiate the numerator and the denominator with respect to rr.\newline Numerator derivative: 2r2r.\newline Denominator derivative: (1/2)(r2+1)1/2(2r)(1/2)(r^2 + 1)^{-1/2}(2r).\newline- Reasoning: L'Hopital's Rule is used to resolve indeterminate forms by differentiating the numerator and denominator.
  4. Simplify after differentiation: Step 44: Simplify the expression after differentiation.\newline- Calculation: limr02r(12)(2rr2+1)\lim_{r\to 0} \frac{2r}{\left(\frac{1}{2}\right)\left(\frac{2r}{\sqrt{r^2 + 1}}\right)}.\newline- Reasoning: Simplifying the derivatives to find the new limit.\newline-
  5. Simplify and evaluate: Step 55: Simplify further and evaluate the limit.\newline- Calculation: limr02rr/r2+1\lim_{r\to 0} \frac{2r}{r/\sqrt{r^2 + 1}}.\newline- Reasoning: Cancel rr from numerator and denominator.\newline-
  6. Final simplification: Step 66: Final simplification and evaluation.\newline- Calculation: limr02r2+1\lim_{r\to 0} 2 \cdot \sqrt{r^2 + 1}.\newline- Reasoning: After canceling rr, the limit depends only on the remaining terms.\newline-

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