Download App

Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}},\text{x}>1$
✓

Answer

$\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\sqrt{\text{x}^2-1}+\sqrt{\text{x}-1}}{\sqrt{\text{x}^2-1}}\times\frac{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}{\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}\times\frac{\sqrt{\text{x}^2-1}}{\sqrt{\text{x}^-1}}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big[\big(\text{x}^2-1\big)-(\text{x}-1)\big]\times\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\big(\text{x}^2-\text{x}\big)\sqrt{\text{x}^2-1}}{\big(\text{x}^2-1\big)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}(\text{x}-1)\sqrt{\text{x}^2-1}}{(\text{x}-1)(\text{x}+1)\big(\sqrt{\text{x}^2-1}-\sqrt{\text{x}-1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}{(\text{x}+1)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}-1\big)}$ $=\frac{\sqrt{2}}{2\big(\sqrt{2}-1\big)}$ $=\frac{\sqrt{2}}{2\times\big(\sqrt{2}-1\big)}\times\frac{\sqrt{2}+1}{\sqrt{2}+1}$ $=\frac{\sqrt{2}+1}{\sqrt{2}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "(Each question 4 marks)" from Limits→Limits — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Solve the following equations: $\sin\text{x}\ \tan\text{x}-1\tan\text{x}-\sin\text{x}$
→
Prove the following identities: $(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2=\sec^2\alpha\sec^2\beta$
→
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow\infty}\Big\{\frac{\text{x}^2+2\text{x}+3}{2\text{x}^2+\text{x}+5}\Big\}^{\frac{3\text{x}-2}{3\text{x}+2}}$
→
Sum the following series to n terms: $1 + 3 + 6 + 10 + 15 + .....$
→
If $\sin\alpha=\frac{4}{5}$ and $\cos\beta=\frac{5}{13},$ prove that $\cos\frac{\alpha-\beta}{2}=\frac{8}{\sqrt{65}}$
→
If the line segment joining the points $P(x_1, y_1)$ and $Q(x_2, y_2)$ subtends an angle $\alpha$ at the origin O, prove that: $\text{O}\text{P}.\text{O}\text{Q}\cos\alpha =\text{x}_1\text{x}_2+\text{y}_1\text{y}_2. $
→
Prove that: $\sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt3}{8}$
→
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{3}+\sqrt{2}\big)}{\text{x}^2-6}$
→
Show that the two formula for the standard deviation of ungrouped data $\sigma=\sqrt{\frac{1}{\text{n}\sum(\text{x}_\text{i}-\overline{\text{X}})^2}}\text{and}\ \sigma^{'}=\sqrt{\frac{1}{\text{n}}\sum{\text{x}_\text{i}}^2-\overline{\text{X}^2}}$ are equivalent, where $ \overline{\text{X}}=\frac{1}{\text{n}}\sum\text{x}_\text{i}.$
→
Let $A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}$. Verify the following identities: $\text{A}\cap(\text{B}\Delta\text{C})=(\text{A}\cap\text{B})\Delta(\text{A}\cap\text{C}).$
→