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        Please help me with the question given in the image attachment
one year ago

Samyak Jain
333 Points
							Put x $\dpi{80} \approx$ 0 in the limit and you will find that it is of the form 1$\dpi{80} \infty$.lim   {f(x)}g(x)  =  e^(x$\dpi{80} \rightarrow$alim{f(x) – 1}.g(x))x$\dpi{80} \rightarrow$aSo, lim [tan($\dpi{80} \pi$/4 + x)]1/x =e^ lim [tan($\dpi{80} \pi$/4 + x) – 1].1/x = e^ lim [{(tan($\dpi{80} \pi$/4) + tanx) / (1 – tan($\dpi{80} \pi$/4)tanx)} – 1] / x     x$\dpi{80} \rightarrow$0                                                                       x$\dpi{80} \rightarrow$0= e^ lim (1 + tanx – 1 + tanx)/x(1 – tanx) = e^ lim 2tanx / x(1 – tanx) = e^ 2 lim (tanx / x) / lim (1 – tanx)      x$\dpi{80} \rightarrow$0                                                x$\dpi{80} \rightarrow$0                                    x$\dpi{80} \rightarrow$0               x$\dpi{80} \rightarrow$0 = e^ 2.1/(1 – 0)  = e2 .Here use lim tanx / x  =  1  and  tan(A+B) = (tanA + tanB) / (1 – tanAtanB)             x$\dpi{80} \rightarrow$0

one year ago
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• 731 Video Lectures
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions